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J.L. HUMAR

The key assets of the book include comprehensive coverage of both the traditional and state-of-the-art numerical techniques of response analysis, such as the analysis by numerical integration of the equations of motion and analysis through frequency domain. The large number of illustrative examples and exercise problems are of great assistance in improving clarity and enhancing reader comprehension. The text aims to benefit students and engineers in the civil, mechanical and aerospace sectors.

DYNAMICS OF STRUCTURES

This major textbook provides comprehensive coverage of the analytical tools required to determine the dynamic response of structures. The topics covered include: formulation of the equations of motion for single- as well as multi-degree-of-freedom discrete systems using the principles of both vector mechanics and analytical mechanics; free vibration response; determination of frequencies and mode shapes; forced vibration response to harmonic and general forcing functions; dynamic analysis of continuous systems; and wave propagation analysis.

THIRD EDITION

DYNAMICS OF STRUCTURES THIRD EDITION JAGMOHAN L. HUMAR

an informa business

Dynamics of Structures Third Edition

Dedicated to Yash and to

Rachna and Rajesh Abhinav and Priya Atul and Deepali and their children who are a source of immense joy to us

Dynamics of Structures Third Edition

Jagmohan L. Humar Carleton University, Ottawa, Canada

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2012 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Version Date: 20120215 International Standard Book Number-13: 978-0-203-11256-4 (eBook - PDF) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http:// www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

Contents

Preface Preface to Second Edition List of symbols 1

Introduction 1.1 Objectives of the study of structural dynamics 1.2 Importance of vibration analysis 1.3 Nature of exciting forces 1.3.1 Dynamic forces caused by rotating machinery 1.3.2 Wind loads 1.3.3 Blast loads 1.3.4 Dynamic forces caused by earthquakes 1.3.5 Periodic and nonperiodic loads 1.3.6 Deterministic and nondeterministic loads 1.4 Mathematical modeling of dynamic systems 1.5 Systems of units 1.6 Organization of the text

xv xvii xxi 1 1 2 3 3 4 4 5 7 8 9 11 12

PART 1 2

Formulation of the equations of motion: Single-degree-of-freedom systems 2.1 Introduction 2.2 Inertia forces 2.3 Resultants of inertia forces on a rigid body 2.4 Spring forces 2.5 Damping forces 2.6 Principle of virtual displacement 2.7 Formulation of the equations of motion 2.7.1 Systems with localized mass and localized stiffness 2.7.2 Systems with localized mass but distributed stiffness 2.7.3 Systems with distributed mass but localized stiffness 2.7.4 Systems with distributed stiffness and distributed mass 2.8 Modeling of multi-degree-of-freedom discrete parameter system 2.9 Effect of gravity load

19 19 19 21 26 29 31 35 35 37 38 42 51 53

vi Contents

2.10 Axial force effect 2.11 Effect of support motion Selected readings Problems 3

4

57 62 63 63

Formulation of the equations of motion: Multi-degree-of-freedom systems 3.1 Introduction 3.2 Principal forces in multi-degree-of-freedom dynamic system 3.2.1 Inertia forces 3.2.2 Forces arising due to elasticity 3.2.3 Damping forces 3.2.4 Axial force effects 3.3 Formulation of the equations of motion 3.3.1 Systems with localized mass and localized stiffness 3.3.2 Systems with localized mass but distributed stiffness 3.3.3 Systems with distributed mass but localized stiffness 3.3.4 Systems with distributed mass and distributed stiffness 3.4 Transformation of coordinates 3.5 Static condensation of stiffness matrix 3.6 Application of Ritz method to discrete systems Selected readings Problems

69 69 71 71 74 76 78 79 80 81 83 89 102 106 109 112 112

Principles of analytical mechanics 4.1 Introduction 4.2 Generalized coordinates 4.3 Constraints 4.4 Virtual work 4.5 Generalized forces 4.6 Conservative forces and potential energy 4.7 Work function 4.8 Lagrangian multipliers 4.9 Virtual work equation for dynamical systems 4.10 Hamilton’s equation 4.11 Lagrange’s equation 4.12 Constraint conditions and Lagrangian multipliers 4.13 Lagrange’s equations for multi-degree-of-freedom systems 4.14 Rayleigh’s dissipation function Selected readings Problems

119 119 119 124 127 132 137 142 145 148 154 155 162 163 165 168 168

PART 2 5

Free vibration response: Single-degree-of-freedom system 5.1 Introduction 5.2 Undamped free vibration 5.2.1 Phase plane diagram

175 175 175 177

Contents vii

5.3

Free vibrations with viscous damping 5.3.1 Critically damped system 5.3.2 Overdamped system 5.3.3 Underdamped system 5.3.4 Phase plane diagram 5.3.5 Logarithmic decrement 5.4 Damped free vibration with hysteretic damping 5.5 Damped free vibration with coulomb damping 5.5.1 Phase plane representation of vibrations under Coulomb damping Selected readings Problems

186 186 188 189 191 192 197 199

6

Forced harmonic vibrations: Single-degree-of-freedom system 6.1 Introduction 6.2 Procedures for the solution of the forced vibration equation 6.3 Undamped harmonic vibration 6.4 Resonant response of an undamped system 6.5 Damped harmonic vibration 6.6 Complex frequency response 6.7 Resonant response of a damped system 6.8 Rotating unbalanced force 6.9 Transmitted motion due to support movement 6.10 Transmissibility and vibration isolation 6.11 Vibration measuring instruments 6.11.1 Measurement of support acceleration 6.11.2 Measurement of support displacement 6.12 Energy dissipated in viscous damping 6.13 Hysteretic damping 6.14 Complex stiffness 6.15 Coulomb damping 6.16 Measurement of damping 6.16.1 Free vibration decay 6.16.2 Forced-vibration response Selected readings Problems

211 211 212 214 218 219 232 237 239 244 249 253 253 255 258 260 265 265 268 268 269 275 275

7

Response to general dynamic loading and transient response 7.1 Introduction 7.2 Response to an Impulsive Force 7.3 Response to general dynamic loading 7.4 Response to a step function load 7.5 Response to a ramp function load 7.6 Response to a step function load with rise time 7.7 Response to shock loading 7.7.1 Rectangular pulse 7.7.2 Triangular pulse

281 281 281 283 284 287 288 293 293 297

202 205 205

viii Contents

7.7.3 7.7.4 7.7.5

Sinusoidal pulse Effect of viscous damping Approximate response analysis for short-duration pulses 7.8 Response to ground motion 7.8.1 Response to a short-duration ground motion pulse 7.9 Analysis of response by the phase plane diagram Selected readings Problems 8

Analysis of single-degree-of-freedom systems: Approximate and numerical methods 8.1 Introduction 8.2 Conservation of energy 8.3 Application of Rayleigh method to multi-degree-of-freedom systems 8.3.1 Flexural vibrations of a beam 8.4 Improved Rayleigh method 8.5 Selection of an appropriate vibration shape 8.6 Systems with distributed mass and stiffness: analysis of internal forces 8.7 Numerical evaluation of Duhamel’s integral 8.7.1 Rectangular summation 8.7.2 Trapezoidal method 8.7.3 Simpson’s method 8.8 Direct integration of the equations of motion 8.9 Integration based on piece-wise linear representation of the excitation 8.10 Derivation of general formulas 8.11 Constant-acceleration method 8.12 Newmark’s β method 8.12.1 Average acceleration method 8.12.2 Linear acceleration method 8.13 Wilson-θ method 8.14 Methods based on difference expressions 8.14.1 Central difference method 8.14.2 Houbolt’s method 8.15 Errors involved in numerical integration 8.16 Stability of the integration method 8.16.1 Newmark’s β method 8.16.2 Wilson-θ method 8.16.3 Central difference method 8.16.4 Houbolt’s method 8.17 Selection of a numerical integration method 8.18 Selection of time step Selected readings Problems

301 304 306 307 313 315 317 317

323 323 325 330 335 339 345 349 352 353 354 355 359 360 364 365 368 370 372 375 377 377 380 381 382 384 387 390 390 390 393 394 395

Contents ix

9

Analysis of response in the frequency domain 9.1 Transform methods of analysis 9.2 Fourier series representation of a periodic function 9.3 Response to a periodically applied load 9.4 Exponential form of Fourier series 9.5 Complex frequency response function 9.6 Fourier integral representation of a nonperiodic load 9.7 Response to a nonperiodic load 9.8 Convolution integral and convolution theorem 9.9 Discrete Fourier transform 9.10 Discrete convolution and discrete convolution theorem 9.11 Comparison of continuous and discrete fourier transforms 9.12 Application of discrete inverse transform 9.13 Comparison between continuous and discrete convolution 9.14 Discrete convolution of an infinite- and a finite-duration waveform 9.15 Corrective response superposition methods 9.15.1 Corrective transient response based on initial conditions 9.15.2 Corrective periodic response based on initial conditions 9.15.3 Corrective responses obtained from a pair of force pulses 9.16 Exponential window method 9.17 The fast Fourier transform 9.18 Theoretical background to fast Fourier transform 9.19 Computing speed of FFT convolution Selected readings Problems

399 399 400 402 405 407 408 410 411 413 416 419 426 432 437 442 444 448 456 459 464 465 469 469 470

PART 3 10

Free vibration response: Multi-degree-of-freedom system 10.1 Introduction 10.2 Standard eigenvalue problem 10.3 Linearized eigenvalue problem and its properties 10.4 Expansion theorem 10.5 Rayleigh quotient 10.6 Solution of the undamped free vibration problem 10.7 Mode superposition analysis of free-vibration response 10.8 Solution of the damped free-vibration problem 10.9 Additional orthogonality conditions 10.10 Damping orthogonality Selected readings Problems

477 477 478 479 483 484 488 490 496 506 509 518 519

11

Numerical solution of the eigenproblem 11.1 Introduction 11.2 Properties of standard eigenvalues and eigenvectors

523 523 524

x

Contents

11.3

Transformation of a linearized eigenvalue problem to the standard form 11.4 Transformation methods 11.4.1 Jacobi diagonalization 11.4.2 Householder’s transformation 11.4.3 QR transformation 11.5 Iteration methods 11.5.1 Vector iteration 11.5.2 Inverse vector iteration 11.5.3 Vector iteration with shifts 11.5.4 Subspace iteration 11.5.5 Lanczos iteration 11.6 Determinant search method 11.7 Numerical solution of complex eigenvalue problem 11.7.1 Eigenvalue problem and the orthogonality relationship 11.7.2 Matrix iteration for determining the complex eigenvalues 11.8 Semidefinite or unrestrained systems 11.8.1 Characteristics of an unrestrained system 11.8.2 Eigenvalue solution of a semidefinite system 11.9 Selection of a method for the determination of eigenvalues Selected readings Problems 12

Forced dynamic response: Multi-degree-of-freedom systems 12.1 Introduction 12.2 Normal coordinate transformation 12.3 Summary of mode superposition method 12.4 Complex frequency response 12.5 Vibration absorbers 12.6 Effect of support excitation 12.7 Forced vibration of unrestrained system Selected readings Problems

13 Analysis of multi-degree-of-freedom systems: Approximate and numerical methods 13.1 Introduction 13.2 Rayleigh–Ritz method 13.3 Application of Ritz method to forced vibration response 13.3.1 Mode superposition method 13.3.2 Mode acceleration method 13.3.3 Static condensation and Guyan’s reduction 13.3.4 Load-dependent Ritz vectors 13.3.5 Application of lanczos vectors in the transformation of the equations of motion

526 527 529 534 538 542 543 546 556 562 564 571 576 576 579 586 586 587 595 596 597 601 601 601 604 608 615 616 626 631 631

635 635 636 653 654 658 663 668 676

Contents xi

13.4

Direct integration of the equations of motion 13.4.1 Explicit integration schemes 13.4.2 Implicit integration schemes 13.4.3 Mixed methods in direct integration 13.5 Analysis in the frequency domain 13.5.1 Frequency analysis of systems with classical mode shapes 13.5.2 Frequency analysis of systems without classical mode shapes Selected readings Problems

679 681 685 694 702 702 707 712 713

PART 4 14

15

Formulation of the equations of motion: Continuous systems 14.1 Introduction 14.2 Transverse vibrations of a beam 14.3 Transverse vibrations of a beam: variational formulation 14.4 Effect of damping resistance on transverse vibrations of a beam 14.5 Effect of shear deformation and rotatory inertia on the flexural vibrations of a beam 14.6 Axial vibrations of a bar 14.7 Torsional vibrations of a bar 14.8 Transverse vibrations of a string 14.9 Transverse vibrations of a shear beam 14.10 Transverse vibrations of a beam excited by support motion 14.11 Effect of axial force on transverse vibrations of a beam Selected readings Problems

719 719 720 722 729

Continuous systems: Free vibration response 15.1 Introduction 15.2 Eigenvalue problem for the transverse vibrations of a beam 15.3 General eigenvalue problem for a continuous system 15.3.1 Definition of the eigenvalue problem 15.3.2 Self-adjointness of operators in the eigenvalue problem 15.3.3 Orthogonality of eigenfunctions 15.3.4 Positive and positive definite operators 15.4 Expansion theorem 15.5 Frequencies and mode shapes for lateral vibrations of a beam 15.5.1 Simply supported beam 15.5.2 Uniform cantilever beam 15.5.3 Uniform beam clamped at both ends 15.5.4 Uniform beam with both ends free 15.6 Effect of shear deformation and rotatory inertia on the frequencies of flexural vibrations 15.7 Frequencies and mode shapes for the axial vibrations of a bar

753 753 754 757 757 759 760 761 762 763 763 766 767 768

731 734 736 738 739 742 746 748 749

772 774

xii Contents

16

15.7.1 Axial vibrations of a clamped–free bar 15.7.2 Axial vibrations of a free–free bar 15.8 Frequencies and mode shapes for the transverse vibration of a string 15.8.1 Vibrations of a string tied at both ends 15.9 Boundary conditions containing the eigenvalue 15.10 Free-vibration response of a continuous system 15.11 Undamped free transverse vibrations of a beam 15.12 Damped free transverse vibrations of a beam Selected readings Problems

776 777

Continuous systems: Forced-vibration response 16.1 Introduction 16.2 Normal coordinate transformation: general case of an undamped system 16.3 Forced lateral vibration of a beam 16.4 Transverse vibrations of a beam under traveling load 16.5 Forced axial vibrations of a uniform bar 16.6 Normal coordinate transformation, damped case Selected readings Problems

799 799

17 Wave propagation analysis 17.1 Introduction 17.2 The Phenomenon of wave propagation 17.3 Harmonic waves 17.4 One dimensional wave equation and its solution 17.5 Propagation of waves in systems of finite extent 17.6 Reflection and refraction of waves at a discontinuity in the system properties 17.7 Characteristics of the wave equation 17.8 Wave dispersion Selected readings Problems

785 786 787 792 794 796 797 798

800 803 805 809 819 825 825 827 827 828 830 833 839 847 851 855 860 860

PART 5 18

Finite element method 18.1 Introduction 18.2 Formulation of the finite element equations 18.3 Selection of shape functions 18.4 Advantages of the finite element method 18.5 Element Shapes 18.5.1 One-dimensional elements 18.5.2 Two-dimensional elements 18.6 One-dimensional bar element

865 865 866 869 870 870 870 871 872

Contents

18.7

19

xiii

Flexural vibrations of a beam 18.7.1 Stiffness matrix of a beam element 18.7.2 Mass matrix of a beam element 18.7.3 Nodal applied force vector for a beam element 18.7.4 Geometric stiffness matrix for a beam element 18.7.5 Simultaneous axial and lateral vibrations 18.8 Stress-strain relationships for a continuum 18.8.1 Plane stress 18.8.2 Plane strain 18.9 Triangular element in plane stress and plane strain 18.10 Natural coordinates 18.10.1 Natural coordinate formulation for a uniaxial bar element 18.10.2 Natural coordinate formulation for a constant strain triangle 18.10.3 Natural coordinate formulation for a linear strain triangle Selected readings Problems

880 883 884 886 886 887 900 902 903 904 911

Component mode synthesis 19.1 Introduction 19.2 Fixed interface methods 19.2.1 Fixed interface normal modes 19.2.2 Constraint modes 19.2.3 Transformation of coordinates 19.2.4 Illustrative example 19.3 Free interface method 19.3.1 Free interface normal modes 19.3.2 Attachment modes 19.3.3 Inertia relief attachment modes 19.3.4 Residual flexibility attachment modes 19.3.5 Transformation of coordinates 19.3.6 Illustrative example 19.4 Hybrid method 19.4.1 Experimental determination of modal parameters 19.4.2 Experimental determination of the static constraint modes 19.4.3 Component modes and transformation of component matrices 19.4.4 Illustrative example Selected readings Problems

931 931 932 932 933 933 933 940 941 941 942 943 944 945 951 952

20 Analysis of nonlinear response 20.1 Introduction 20.2 Single-degree-of freedom system

911 915 921 926 926

957 960 961 971 972 975 975 977

xiv Contents

20.2.1 Central difference method 20.2.2 Newmark’s β Method 20.3 Errors involved in numerical integration of nonlinear systems 20.4 Multiple degree-of-freedom system 20.4.1 Explicit integration 20.4.2 Implicit integration 20.4.3 Iterations within a time step Selected readings Problems Answers to selected problems Index

979 981 985 990 990 995 999 1000 1000 1003 1019

Preface

Since the publication of its first edition in 1990, this book continues to attract the interest of students, academics, and professionals. It has been acquired by prominent engineering libraries around the world and is used as a text book in many educational institutions. This has encouraged the author to write this third edition. As in the case of its predecessors, the motivation for this edition of the book is to help engineers and scientists acquire an understanding of the dynamic response of structures and of the analytical tools required for determining such response. The book should be equally helpful to persons working in the field of civil, mechanical or aerospace engineering. The major revisions in this edition comprise the addition of three new chapters: Chapter 18 on Finite Element Method, Chapter 19 on Component Mode Synthesis, and Chapter 20 on the Analysis of Nonlinear Response. Sections dealing with finite element method that were previously in Chapter 3 have been moved to the new Chapter 18, which now includes significant additional material on the subject. Chapter 19 is completely new. It deals with the important topic of component mode synthesis, which should be of equal interest to civil, mechanical and aerospace engineers. Chapter 20 on the Analysis of Nonlinear Response now includes material that was previously scattered in Chapters 8 and 13. It also contains significant additional material. The new chapters also contain new end of chapter exercises. Solutions to the end-of-chapter problems are available to instructors as a download from the publisher’s web site. A limited number of print copies may also be available. The contents of the Preface to the Second Edition are relevant to this edition too. That preface provides details of the topics covered and the organization of the of the material included. Here, as in the previous editions, the material included in this book has been drawn from the vast wealth of available information. It is difficult to acknowledge the sources for all of the information provided. The author again offers his apologies to all researchers who have not been adequately recognized. The author wishes to acknowledge the contribution made by his many students and colleagues in the preparation of this book.

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Preface to Second Edition

As in the case of its predecessor, the motivation for this second edition of the book is to help engineers and scientists acquire an understanding of the dynamic response of structures and of the analytical tools required for determining such response. The book should be equally helpful to persons working in the field of civil, mechanical or aerospace engineering. For the proper understanding of an analytical concept, it is useful to develop an appreciation of the mathematical basis. Such appreciation need not depend on a rigorous treatment of the subject matter; a physical understanding of the concepts is in most cases adequate, and perhaps more meaningful to an engineer. The book attempts to explain the mathematical basis for the concepts presented, mostly in physically motivated terms or through heuristic argument. No special mathematical background is required of the reader, except for a basic knowledge of college algebra and calculus and engineering mechanics. The essential steps in the dynamic analysis of a system are: (a) mathematical modeling (b) formulation of the equations of motion, and (c) solution of the equations. Modeling techniques can be divided into two broad categories. In one technique, the system is modeled as an assembly of rigid body masses and massless deformable elements. Systems modeled in this manner are referred to as discrete parameter systems. In the other technique of modeling, both mass and deformability are assumed to be distributed throughout the extent of the system which is treated as continuous. Systems modeled in this manner are called continuous or distributed parameter systems. In general, a continuous model will better represent the behavior of a dynamical system. However, in most practical situations, the equations of motion of a continuous system are too difficult or impossible to solve. Therefore, in a majority of cases, dynamic analysis of engineering structures must rely on a representation of the structure by a discrete parameter model. The contents of the book reflect this emphasis on the use of discrete models. The first three parts of the book are devoted to the analysis of response of discrete systems. Part I consisting of Chapters 2 through 4 deals with the formulation of equations of motion of discrete parameter systems. However, the methods of analytical mechanics presented in Chapter 4 are, equally applicable to continuous system. Examples of such applications are presented later in the book. Part II of the book covering Chapters 5 through 9 deals with the solution of equation of motion for a single-degree-of-freedom system. Part III consisting of Chapters 10 through 13 discusses the solution of equations of motion for multi degree-of-freedom systems.

xviii Preface to Second Edition

Part IV of the book covering Chapters 14 through 17 is devoted to the analysis of continuous system. Again, the subject matter is organized so that the formulation of equations of motion is presented first followed by a discussion of the solution techniques. The book is organized so as to follow the logical succession of steps involved in the analysis. Many readers may prefer to complete a study of the single degree-offreedom systems, from formulation of equation to their solution, before embarking on a study of multi degree-of-freedom systems. This can be easily achieved by selective reading. The book chapters have been planned so that Chapters 3 and 4 relating to the formulation of equations of motion of a general system need not be studied prior to studying the material in Chapters 5 through 9 on the solution of equations of motion for a single degree-of-freedom system. A development that has had a profound effect in the recent times on procedures for the analysis of engineering systems is the advent of digital computers. The ability of computers to manage vast amounts of information and the incredible speed with which they can process numerical data has shifted the emphasis from closed from solutions and approximate methods suitable for hand computations to solution of discrete models and numerical techniques of analysis. At the same time, computers have allowed the routine solution of problems vastly greater in size and complexity than was possible only a decade or two ago. The emphasis on discrete methods and numerical solutions is reflected in the contents of the present book. Chapter 8 on single degree-of-freedom systems and Chapter 13 on multi degree-of-freedom systems are devoted exclusively to numerical techniques of solution. A fairly detailed treatment of the frequency domain analysis is included in Chapters 9 and 13 in recognition of the efficiency of this technique in the numeric computation of response. Also, a detailed treatment of the solution of discrete eigenproblems which plays a central role in the numerical analysis of response is included in Chapter 11. It is recognized that the field of computer hardware as well as software is undergoing revolutionary development. The continuing evolution of personal computers with vastly improved processing speeds and memory capacity and the ongoing development of new programming languages and software tools means that algorithms and programming styles must continue to change to take advantage of the progress made. Program listings or detailed algorithms have not therefore been included in the book. The author believes that in a book like this, it is more useful to provide the necessary background material for an appreciation of the physical behavior and the analytical concepts involved as well as to present the development of methods that are suitable to numeric computations. Hopefully, this will give enough information for the reader to be able to develop his/her own algorithms or to make an informed and intelligent use of existing software. The material included in this book has been drawn from the vast wealth of available information. Some of it has now become a part of the historical development of structural dynamics, other is more recent. It is difficult to acknowledge the sources for all of the information provided. The author offers his apologies to all researchers who have not been adequately recognized. References have been omitted from the text to avoid distracting the reader. However, where appropriate, a brief list of suitable material for further reading is provided at the end of each chapter.

Preface to Second Edition xix

The style of presentation and the emphasis are author’s own. The contents of the book have been influenced by author’s experience in teaching and research and by the research studies carried out by him and his students. A large number of examples have been included in the text; since they provide the most effective means of developing an understanding of the concepts involved. Exercise problems have also been included at the end of each chapter. They will provide the reader useful practice in the application of techniques presented. In preparing this second edition, the errors that had inadvertently crept into the first edition have been corrected. The author is indebted to all those readers who brought such errors to his attention. Several sections of the book have been revised and some new concepts and analytical techniques have been included to make the book as comprehensive as possible within the boundary of its scope. Also included are additional end-of-chapter exercises for the benefit of the reader. The author wishes to acknowledge the contribution made by his many students and colleagues in the preparation of this book.

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List of symbols

The principal symbols used in the text are listed below. All symbols, including those listed here, are defined at appropriate places within the text, usually at the time of their first occurrence. Occasionally, the same symbol may be used to represent more than one parameter, but the meaning should be quite unambiguous when read in context. Throughout the text, matrices are represented by bold face upper case letters while vectors are generally represented by bold face lower case letters An overdot signifies differential with respect to time and a prime stands for differentiation with respect to the argument of the function a a an aij am A Aa Ad Av A ˜ A b bn b bm B B c ccr cg cn cs cij c¯ c∗ c

acceleration; constant; linear dimension decay parameter in exponential window method coefficient of Fourier series cosine term flexibility influence coefficient real part of mth eigenvector constant; cross-sectional area amplitude of dynamic load factor for acceleration amplitude of dynamic load factor for displacement amplitude of dynamic load factor for velocity amplification matrix; flexibility matrix; square matrix transformed square matrix constant; linear dimension; width of beam cross section coefficient of Fourier series sine term vector of body forces per unit volume imaginary part of mth eigenvector constant; differential operator square matrix damping constant; velocity of wave propagation critical damping constant velocity of wave group coefficient of Fourier series term, constant internal damping constant damping influence coefficient damping constant per unit length generalized damping constant vector of weighting factors in expansion theorem

xxii

List of symbols

C Cn C ˜ C∗ , C d dn D D e E Em EA EI E f f (x) fd fD fG fI fS fSt f0 f f fD fG fI fIi fS fSi F Fx , Fy , Fz F Fa Fc g g(t) gˆ G G1 , G2 G() ˆ G() GJ G Gd Gf

constant modal damping constant for the nth mode damping matrix; transformation matrix transformed damping matrix diameter constant dynamic load factor diagonal matrix; dynamic matrix; elasticity matrix eccentricity of unbalanced mass modulus of elasticity remainder term in numerical integration formula axial rigidity flexural rigidity dynamic matrix = D−1 undamped natural frequency in cycles per sec eigenfunction of a continuous system damped natural frequency damping force force due to geometric instability inertia force spring force total of spring force and damping force for hysteretic damping frequency of applied load in cycles per sec vector representing spatial variation of exciting force vector of forces acting on element nodes vector of damping forces vector of geometric instability forces vector of inertia forces; vector of global inertia forces vector of inertia forces in element i vector of spring forces; vector of global spring forces vector of spring forces in element i force components of force vector along Cartesian coordinates force vector vector of applied forces vector of constraint forces acceleration due to gravity forcing function scaled forcing function e−at g(t) constant; modulus of rigidity constants Fourier transform of g(t) Fourier transform of g(t) ˆ torsional rigidity flexibility matrix residual flexibility matrix inertia relief flexibility matrix

List of symbols

Gk h h(t) ¯ h(t) ˆh(t) H(ω0 ), H() ¯ H() ˆ H(t) H i i I IA I0 I j j J k kG kT kij k k¯ k∗ k K Kn K Ke ˆi K KG ˜ K∗ , K ˜ cc , K ˜ ss K l L L LK LM m m0 mij m ¯ m∗ M

flexibility matrix of retained modes height; time interval unit impulse response periodic unit impulse response scaled unit impulse function h(t)e−at complex frequency response, Fourier transform of h(t) ¯ periodic complex frequency response, Fourier transform of h(t) ˆ Fourier transform of h(t) matrix of frequency response functions imaginary number; integer unit vector along x axis impulse; moment of inertia mass moment of inertia for rotation above point A functional; mass moment of inertia for rotation about the mass center identity matrix integer unit vector along y axis polar moment of inertia spring constant; stiffness; integer; wave number geometric stiffness tangent stiffness stiffness influence coefficient shape constant for shear deformation spring constant per unit length generalized stiffness unit vector along z axis differential operator modal stiffness for the nth mode stiffness matrix stiffness matrix of an element augmented stiffness matrix for element i geometric stiffness matrix transformed stiffness matrix effective constrained coordinate stiffness matrix length Lagrangian; length; length of an element operator matrix; vector of interpolation functions lower triangular factor of stiffness matrix lower triangular factor of mass matrix integer; mass; mass per unit length mass; unbalanced mass mass influence coefficient mass per unit length; mass per unit area generalized mass concentrated mass, differential operator; moment

xxiii

xxiv

List of symbols

MI Mn Ms M0 M Me ˆi M ˜ M∗ , M n N Ni N p pn p¯ p∗ p(λ) p p pe P PI P0 , p0 P q qi q q q˜ qe qi Q Qi Q r r r(t) r¯(t) r R Ra Rd Rv Ri Ra Rd Rv

inertial moment modal mass for the nth mode moment due to internal damping forces concentrated mass mass matrix mass matrix of an element augmented mass matrix for element i transformed mass matrix integer normal force; number of degrees of freedom interpolation function transformation matrix integer; force modal force in the nth mode force per unit length generalized force characteristic polynomial left eigenvector; force vector; global force vector vector of generalized coordinates equivalent forces at element nodes axial force; concentrated applied load inertial force; inertia relief projection matrix amplitude of applied force matrix of left eigenvectors integer generalized coordinate right eigenvector; global nodal parameters vector of generalized coordinates transformed eigenvector vector of displacements at element nodes nodal parameters for element i applied force generalized force matrix of eigenvectors, orthogonal transformation matrix common ratio; constant; integer; radius of gyration rank of a matrix; radius vector response due to unit initial displacement response due to periodic unit displacement changes vector of applied forces per unit volume Rayleigh dissipation function; reaction; remainder term inertance receptance mobility magnitude of ith corrective force impulse inertance matrix receptance matrix mobility matrix

List of symbols xxv

RS s s¯ s(t) s¯(t) S S Sn t tp t T T Td TR T0 T u(t), u ug ui ux uy u0 ut us u(t) ¯ u U U() U v v(x) v0 v v¯ V V0 V w(x) WD We Wi Ws x x¯ X y

vector of support reactions complex eigenvalue conjugate of complex eigenvalue s response due to initial unit velocity response due to a periodic unit velocity changes axial force matrix of complex eigenvalues; transformation matrix matrix for sweeping the first n eigenvectors time time at peak response surface forces per unit area torque kinetic energy; tensile force; undamped natural period damped natural period transmission ratio period of applied load transformation matrix; tridiagonal matrix displacement ground displacement constrained coordinate; displacement along degree-of-freedom i displacement along x direction displacement along y direction initial displacement absolute displacement static displacement periodic displacement response displacement vector complex frequency response; strain energy Fourier transform of u(t) upper triangular matrix, complex frequency response matrix velocity comparison function initial velocity complex eigenvector conjugate of complex eigenvector potential energy; shear force base shear matrix of complex eigenvectors comparison function energy loss per cycle in viscous damping work done by external forces energy loss per cycle; work done by internal forces work done by elastic force Cartesian coordinate coordinate of the mass center Lanczos transformation matrix Cartesian coordinate

xxvi

List of symbols

y0n y y0 z α αi α β γ δ δ δ(x) δij δu δz δε δθ , δφ δqe δr δu δWe δWi δWei δWS  st t   ε ε η η(t) ηk η θ κ λ  µ µ(t) µm νm ξ ξh ξk ρ

initial value of the nth normal coordinate vector of normal coordinates vector of initial values of the normal coordinates generalized coordinate angular shear deformation; coefficient; constant; parameter generalized coordinate vector of generalized coordinates constant; frequency ratio; parameter angle; inverse eigenvalue; parameter deflection; eigenvalue; eigenvalue measured from a shifted origin logarithmic decrement delta function Kronecker delta vector of virtual displacements in an element virtual displacement vector of virtual displacements in an element virtual rotation vector of virtual displacements at element nodes virtual displacement vector virtual virtual work done by external forces virtual work done by internal forces virtual work done by forces acting on internal elements virtual work done by axial force displacement static deflection increment of time increment of frequency vector of displacements strain; quantity of a small value strain vector; real part of complex eigenvector hysteretic damping constant, angle corrective response imaginary part of eigenvector imaginary part of complex eigenvector angular displacement; flexural rotation; polar coordinate; parameter curvature eigenvalue; Lagrangian multiplier; wave length matrix of eigenvalues coefficient of friction; eigenvalue; eigenvalue shift unit step function real part of mth eigenvalue imaginary part of mth eigenvalue damping ratio; spatial coordinate equivalent hysteretic damping ratio real part of eigenvector root of difference equation, mass per unit volume

List of symbols

ρ ρh ρ(A) σ σD σ τ φ φ φ(x) φh φ  k f r χ ψ ψ(x) a c s ω ωd ω0 ω  ∇

amplitude of motion; Rayleigh quotient amplitude of motion for hysteretic damping spectral radius of A stress damping stress stress vector time angle; normalized eigenvector or mode shape; phase angle potential function; spherical coordinate normalized eigenfunction phase angle for hysteretic damping mode shape, mass-normalized mode shape modal matrix, matrix of mass-normalized mode shapes matrix of retained modes flexible body modes rigid body modes response amplitude shape vector shape function matrix of attachment modes matrix of static constraint modes matrix of static constraint modes undamped natural frequency in rad/s damped natural frequency frequency of applied load in rad/s vector of natural frequencies frequency of the exciting force gradient vector

xxvii

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Chapter 1

Introduction

1.1

OBJECTIVES OF THE STUDY OF STRUCTURAL DYNAMICS

The response of physical objects to dynamic or time-varying loads is an important area of study in physics and engineering. The physical object whose response is sought may either be treated as rigid-body or considered to be deformable. The subject of rigid body dynamics treats the physical objects as rigid bodies that undergo motion without deformation when subjected to dynamic loading. The study of rigid-body motion has many applications, including, for example, the movement of machinery, the flight of an aircraft or a space vehicle, and the motion of earth and the planets. In many instances, however, dynamic response involving deformations rather than simple rigid-body motion is of primary concern. This is particularly so in the design of structures and structural frames that support manufactured objects. Structural frames form a part of a wide variety of physical objects created by human beings: for example, automobiles, ships, aircraft, space vehicles, offshore platforms, buildings, and bridges. All of these objects, and hence the structure supporting them, are subjected to dynamic disturbances during their service life. Dynamic response involving deformations is usually oscillatory in nature, in which the structure vibrates about a configuration of stable equilibrium. Such equilibrium configuration may be static, that is, time invariant, or it may be dynamic involving rigid-body motion. Consider, for example, the vibrations of a building under the action of wind. In the absence of wind, the building structure is in a state of static equilibrium under the loads acting on it, such as those due to gravity, earth pressure, and so on. When subjected to wind, the structure oscillates about the position of static equilibrium as shown in Figure 1.1. An airplane in flight provides an example of oscillatory motion about an equilibrium configuration that involves rigid-body motion. The aircraft can be idealized as consisting of rigid-body masses of fuselage and the engines connected by flexible wing structure (Figure 1.2). When in flight, the whole system moves as a rigid body and may, in addition, be subjected to oscillatory motion transverse to the flight plane. Motions involving deformation are caused by dynamic forces or dynamic disturbances. Dynamic forces may, for example, be induced by rotating machinery, wind, water waves, or a blast. A dynamic disturbance may result from an earthquake during which the motion of the ground is transmitted to the supported structure. Later in this chapter we discuss briefly the nature of some of the dynamic forces and disturbances.

2

Dynamics of structures

Figure 1.1 Oscillatory motion of a building frame under wind load.

Figure 1.2 Aeroplane in flight.

Whatever be the cause of excitation, the resulting oscillatory motion of the structure induces displacements and stresses in the latter. An analysis of these displacements and stresses is the primary objective of a study of the dynamics of structures. 1.2

IMPORTANCE OF VIBRATION ANALYSIS

The analysis of vibration response is of considerable importance in the design of structures that may be subjected to dynamic disturbances. Under certain situations vibrations may cause large displacements and severe stresses in the structure. As we shall see later, this may happen when the frequency of the exciting force coincides with a natural frequency of the structure. Also, fluctuating stresses, even of moderate intensity, may cause material failure through fatigue if the number of repetitions is large enough. Oscillatory motion may at times cause wearing and malfunction of machinery. Also, the transmission of vibrations to connected structures may lead to undesirable results. Vibrations induced by rotating or reciprocating machinery may, for example, be transmitted through the supporting structure to delicate instruments mounted elsewhere on it, causing such instruments to malfunction. Finally, when the structure is designed for human use, vibratory motion may result in severe discomfort to the occupants.

Introduction 3

With progress in engineering design increasing use is being made of lightweight, high strength materials. As a result, modern structures are more susceptible to critical vibrations. This is as true of mechanical structures as of buildings and bridges. Today’s buildings and bridges structures are, for example, lighter, more flexible, and are made of materials that provide much lower energy dissipation, all of which may contribute to more intense vibration response. Dynamic analysis of structures is therefore even more important for modern structures, and this trend is likely to continue. It is apparent from the foregoing discussion that vibrations are undesirable for engineering structures. This is in general true except for certain mechanical machinery which relies on controlled vibration for its functioning. Such machinery includes, for example, vibratory compactors, sieves, vibratory conveyors, certain types of drills, and pneumatic hammers. In any case, whether or not the vibrations arise from natural causes or are induced on purpose, the structure subjected to such vibrations must be designed for the resultant displacements and stresses.

1.3

N ATURE OF EXCITING FORCES

As stated earlier, the dynamic forces acting on a structure may result from one or more of a number of different causes, and it may be useful to categorize these forces according to the source of their origin, such as, for example, rotating machinery, wind, blast, or earthquake. The exciting forces may also be classified according to the nature of their variation with time as periodic, nonperiodic, or random. It is also useful to classify dynamic forces as deterministic, being specified as a definite function of time, or nondeterministic, being known only in a statistical sense. In the following, we discuss briefly each of these classifications.

1.3.1 Dynamic forces caused by rotating machinery Rotating machinery that is not fully balanced about the center of rotation will give rise to exciting forces that vary with time. Consider, for example, a rotating motor that has an eccentric mass m0 attached to it at a distance e from the center of rotation, as shown in Figure 1.3a. If the motor is rotating with a constant angular velocity  rad/s, the centrifugal force acting on the unbalanced mass is em0 2 directed away from center and along the radius connecting the eccentric mass to the center. If time is measured from the instant the radius vector from the center of rotation to the mass is horizontal, the horizontal component of the centrifugal force at time t is given by p(t) = em0 2 cos t

(1.1)

Force p(t) is shown as a function of time in Figure 1.3b. If the supporting table is free to translate in a horizontal direction, force p(t) will cause the table to vibrate in that direction. Dynamic forces arising from unbalanced rotating machinery are quite common in mechanical systems, and the supporting structure must in such cases be designed to withstand the resulting deformations and stresses.

4

Dynamics of structures

Figure 1.3 (a) Rotating machinery with unbalanced mass; (b) horizontal component of centrifugal force.

1.3.2 Wind loads Structures subjected to wind experience aerodynamic forces which may be classified as drag forces, which are parallel to the direction of wind, and lift forces, which are perpendicular to the wind. Both forces depend on the wind velocity, the wind profile along the height of the structure, and the characteristics of the structure. Winds close to the surface of the earth are affected by turbulence and hence vary with time. The response of the structure to the wind is thus a dynamic phenomenon and a precise estimate of the displacements and stresses induced by the wind can be obtained only through a dynamic analysis. For the purpose of design, wind forces are often converted into equivalent static forces. This approach while reasonable for low rise, comparatively stiff structures, may not be appropriate for structures that are tall, light, flexible, and possess low damping resistance. Estimates of design wind speeds are obtained by measurements of wind in an open exposure, often at an airport, at a standard height, usually 10 m or 30 ft. Records are kept of maximum daily time-averaged mean wind speeds. Obviously, the mean wind will depend on the time used for the purpose of averaging. Design codes generally specify the use of a maximum mean wind with a given recurrence period. A typical value of recurrence period for strength design of buildings subjected to wind loads is 30 years. The corresponding design wind is usually obtained by a statistical analysis of the recorded data on hourly mean winds. The variation of wind along the height, called wind profile, is determined on the basis of analytical studies and experimental observation. A similar approach is used to model wind turbulence or the variation of wind with time. The design mean wind speed, the wind profile and the wind turbulence together constitute the input data for a dynamic analysis for wind. It is evident that the effect of wind cannot be represented by a set of forces that are definite functions of time, since the wind loads are known only in a statistical sense.

1.3.3 Blast loads A dynamic load of considerable interest in the design of certain structures is that due to a blast of air striking the structure. The blast or shock wave is usually caused by

Introduction 5

Figure 1.4 Pressure-time curve for a blast.

the detonation of a conventional explosive such as TNT or a bomb. In either case, the explosion results in the rapid release of a large amount of energy. A substantial portion of the energy released is expended in driving a shock wave whose front consists of highly compressed air. The peak overpressure (pressure above atmospheric pressure) in the shock front decreases quite rapidly as the shock wave propagates outward from the center of explosion. The overpressure in a shock wave arriving at a structure will thus depend on both the distance from the center of explosion and the strength of the explosive. The latter is measured in terms of the weight of a standard explosive, usually TNT, required to release the same amount of energy. Thus a 1-kiloton bomb will release the same amount of energy as the detonation of 1000 tons of TNT. Empirical equations derived on the basis of observations are available for estimating the peak overpressure in a blast caused by the detonation of an explosive of given strength and striking a structure located at a given distance from the center of explosion. The overpressure rapidly decreases behind the front, and at some time after the arrival of the shock wave, the pressure may, in fact, become negative. The duration of the positive phase and the variation of the blast pressure during that phase can also be obtained from available empirical equations. In summary, a blast load can be represented by a pressure wave in which the pressure rises very rapidly or almost instantaneously and then drops off fairly rapidly according to a specified pressure–time relationship. A typical blast load history is shown in Figure 1.4.

1.3.4 Dynamic forces caused by earthquakes Ground motions resulting from earthquakes of sufficiently large magnitude are one of the most severe and disastrous dynamic disturbances that affect human-made structures. Earthquakes are believed to result from a fracture in the earth’s crust. The forces that cause such fractures are called tectonic forces. In fact, they are the very forces that have caused the formation of mountains and valleys and the oceans. They arise because of a slow convective motion of the earth’s mantle that underlies the crust. This movement sets up elastic strains in the crustal rock. When the ability of the rock to sustain the elastic strain imposed on it is exceeded, a fracture is initiated at a zone of

6

Dynamics of structures

weakness in the rock. Fracturing relieves the elastic strains, causing the opposite sides of the fault to rebound and slip with respect to each other. The consequent release of the elastic strain energy stored in the rock gives rise to elastic waves which propagate outwards from the source fault. Before arriving at a specific location on the earth’s surface these waves may undergo a series of reflections and refractions. The earthquake wave motion is very complex. The effect of such a motion on the supported structure can best be assessed by obtaining measurements of the time histories of ground displacements or accelerations by means of special measuring instruments called seismographs, and then analyzing the structure for the recorded motion. It is generally more convenient and common to obtain measurements of the ground acceleration. Then if required, the velocity and displacement histories are derived from the recorded acceleration history by a process of successive numerical integration. Figure 1.5 shows the acceleration history recorded at El Centro, California, during an earthquake that took place in May 1940. Velocity and displacement histories obtained by successive integration are also shown. Ground motions induced by an earthquake cause dynamic excitation of a supported structure. As we shall see later, the time-varying support motion can be translated into a set of equivalent dynamic forces that act on the structure and cause it to deform relative to its support. If a ground motion history is specified, it is possible to analyze the structure to obtain estimates of the deformations and stresses induced in it. Such analytical studies play an important role in the design of structures expected to undergo seismic vibrations.

Figure 1.5 Imperial Valley earthquake, El Centro site, May 18, 1940, component N-S.

Introduction 7

1.3.5 Periodic and nonperiodic loads Dynamic loads vary in their magnitude, direction, or position with time. It is, in fact, possible for more than one type of variation to coexist. As an example, earthquake induced forces vary both in magnitude and direction. However, by resolving the earthquake motion into translational components in three orthogonal directions and the corresponding rotational components, the earthquake effect can be defined in terms of six component forces and moments, each of which varies only in the magnitude with time. The constant-magnitude centrifugal force caused by imbalance in a rotating machinery can be viewed as a force of constant magnitude that is continually varying in direction with time. Alternatively, we can interpret the force as consisting of two orthogonal components, each of which varies in magnitude with time. A wheel load rolling along the deck of a bridge provides an instance of a force that varies in its location with time. A special type of dynamic load that varies in magnitude with time is a load that repeats itself at regular intervals. Such a load is called a periodic load. The era of load duration that is repeated is called a cycle of motion and the time taken to complete a cycle is called the period of the applied load. The inverse of the period, that is the number of cycles per second, is known as the frequency of the load. A general type of periodic load is shown in Figure 1.6 which also identifies the period of the load. The harmonic load caused by an unbalanced rotating machine, shown in Figure 1.3b, is a more regular type of periodic load. Loads that do not show any periodicity are called nonperiodic loads. A nonperiodic load may be of a comparatively long duration, A rectangular pulse load imposed on a simply supported beam by the sudden application of a weight that remains in contact with the beam from the instance of its initial application is an example of a long-duration nonperiodic load. Such a load is shown in Figure 1.7. Nonperiodic loads may also be of short duration or transient, such as, for example, an air blast striking a building. When the duration of the transient load is very short, the load is often referred to as an impulsive load. A load or a disturbance that varies in a highly irregular fashion with time is sometimes referred to as a random load or a random disturbance. Ground acceleration resulting from an earthquake provides one example of a random disturbance.

Figure 1.6 General periodic load.

8

Dynamics of structures

Figure 1.7 Simply supported beam subjected to a rectangular pulse load.

1.3.6 Deterministic and nondeterministic loads From the discussion in the previous paragraphs we observe that certain types of loads can be specified as definite functions of time. The time variation may be represented by a regular mathematical function, for example, a harmonic wave, or it may be possible to specify the load only in the form of numerical values at certain regularly spaced intervals of time. Loads that can be specified as definite functions of time, irrespective of whether the time variation is regular or irregular, are called deterministic loads, and the analysis of a structure for the effect of such loads is called deterministic analysis. The harmonic load imposed by unbalanced rotating machinery is an example of a deterministic load that can be specified as a mathematical function of time. A blast load is also a deterministic load, and it may be possible to represent it by a mathematical curve that will closely match the variation. A measured earthquake accelerogram is a deterministic load that can only be specified in the form of numerical values at selected intervals of time. Certain types of loads cannot be specified as definite functions of time because of the inherent uncertainty in their magnitude and the form of their variation with time. Such loads are known in a statistical sense only and are described through certain statistical parameters such as mean value and spectral density. Loads that cannot be described as definite functions of time are known as nondeterministic loads. The analysis of a structure for nondeterministic loads yields response values that are themselves defined only in terms of certain statistical parameters, and is therefore known as nondeterministic analysis. Earthquake loads are, in reality, nondeterministic because the magnitude and frequency distribution of an acceleration record for a possible future earthquake cannot be predicted with certainty but can be estimated only in a probabilistic sense. Wind loads are quite obviously nondeterministic in nature. Throughout this book we assume that the loads are deterministic or can be specified as definite functions of time. The methods of analysis and the resultant response will also therefore be deterministic.

Introduction 9

1.4

MATHEMATICAL MODELING OF DYNAMIC SYSTEMS

In an analysis for the response of a system to loads that are applied statically, we need to concern ourselves with only the applied loads and the internal elastic forces that oppose the former. Dynamic response is much more complicated, because in addition to the elastic forces, we must contend with inertia forces and the forces of damping resistance that oppose the motion. Static response can, in fact, be viewed as a special case of dynamic response in which the accelerations and velocities are so small that the inertia and damping forces are negligible. Before carrying out any analysis, the physical system considered must be represented by a mathematical model that is most appropriate for obtaining the desired response parameters. In either a static or a dynamic analysis of a structure, the response parameters of interest are displacements, and internal forces or stresses. Critical values of these parameters are required in the design. For a static case, the response is a function of one or more spatial coordinates; for a dynamic case the response depends on both the space variables and the time variable. Apparently, the dynamic response must be governed by partial differential equations involving the space and the time coordinates. This is indeed so. However, in many cases it is possible to model the system as an assembly of rigid bodies that have mass but are not deformable or have no compliance, and massless spring-like elements that deform under load and provide the internal elastic forces that oppose such deformation. The response of such a model is completely defined by specifying the displacements along certain coordinates that determine the position of the rigid-body masses in space. A system modeled as above is referred to as a discrete system or a discrete parameter system. The number of coordinate directions along which values of the response parameters must be specified in order to determine the behavior of the mathematical model completely is called the number of degrees of freedom. The response of a discrete parameter system is governed by a set of ordinary differential equations whose number is equal to the number of degrees of freedom. As an example of the modeling process, consider a bar clamped at its left-hand end and free at the other, as shown in Figure 1.8a. Let the cross-sectional dimensions of the bar be small as compared to its length, and let the bar be constrained so that its fibers can move in only an axial direction. A cross section such as AA will move in the positive and negative directions of the x axis, and the position of the cross section at a time t will be a function of both the spatial location of the cross section, that is, the value of x, and the value of time t. The axial vibrations of the bar are thus governed by a partial differential equation involving the independent variables x and t. It may, however, be reasonable to model this bar by the assembly of a series of rigid masses and interconnecting springs as shown in Figure 1.8b. The selection of the number of mass elements will depend upon the accuracy that is desired. Supposing that this number is N, we must determine the horizontal displacement of each of the N masses to describe completely the response of the system at a given time t. The system is therefore said to have N degrees of freedom, and its response is governed by N ordinary differential equations. For the bar model just described, the N degrees of freedom or coordinates correspond to displacements along a Cartesian direction. Alternative choices for coordinates are possible. In fact, in many situations coordinates known as generalized coordinates

10

Dynamics of structures

may prove to be more effective. We discuss the meaning and application of such coordinates in the subsequent chapters of the book. For a large majority of physical systems, discrete models consisting of an assembly of rigid mass elements and flexible massless elements are quite adequate for the purpose of obtaining the dynamic response. It should, however, be recognized that, in general, discrete modeling is an idealization because all mass elements will possess certain compliance and all flexible elements will possess some mass. In fact, in certain situations, a model in which both the mass and the flexibility are distributed may be better able to represent the physical system under consideration. Such a model is referred to as a continuous system or a distributed parameter system. Its response is governed by one or more partial differential equations. In general, the analysis of a discrete system is much simpler than that of a continuous system. Furthermore, it is usually possible to improve the accuracy of the results obtained from the analysis of a discrete model by increasing the number of degrees of freedom in the model. Discrete modeling is therefore usually the preferred approach in the dynamic analysis of structures. Consequently, a major portion of this book is devoted to the modeling and analysis of discrete systems. Sufficient information is, however, provided on the analysis of simple continuous systems as well. In the dynamic analysis of engineering structures, it is generally assumed that the characteristics of the system, that is, its mass, stiffness, and damping properties, do not vary with time. It is further assumed that deformations of the structure are small and that the deforming material follows a linear stress–strain relationship. When the foregoing assumptions are made, the structure being analyzed is said to be linear and the principle of superposition is valid. This principle implies that if u1 is the response of the structure to an applied load p1 and u2 is the response to another load p2 , the total response of the structure under simultaneous application of p1 and p2 is obtained by summing the individual responses u1 and u2 so that u = u1 + u2

Figure 1.8 (a) Bar undergoing axial vibrations; (b) lumped mass model of the bar in (a).

(1.2)

Introduction 11

The principle of superposition usually results in considerable simplification of the analysis. Fortunately, for a majority of engineering structures in service, it is quite reasonable to assume that the principle of superposition is valid and can be applied to their analysis. In a few situations, however, structural deformations may be quite large. Also, the stress–strain relationships may become nonlinear as the material deforms. Both of these conditions will introduce nonlinearity in the system. The nonlinearity associated with large deformations is called geometric nonlinearity, and that associated with nonlinear stress–strain relationship is called material nonlinearity. These nonlinearities, particularly the material nonlinearity, may have to be taken into account in analyzing structures that are strained into the postelastic range, which may, for example, be the case under a severe earthquake excitation. A majority of analysis procedures described in this book assume that the structure is linear. However, a brief discussion of nonlinear analysis procedures is presented in Chapter 20. 1.5

SYSTEMS OF UNITS

Two different systems of units are in common use in engineering practice. One of these is the International System of metric units, commonly referred to as the SI units. The other system is the system of Imperial units. In the International System, the basic unit of length is the meter (m), the basic unit of mass is the kilogram (kg), and the basic unit of time is the second (s). The unit of force is a derived unit and is known as a newton. A newton is defined as a force that will produce an acceleration of 1 m/s2 on a mass of 1 kg. Since, according to the Newton’s law of motion, force is equal to the product of mass and acceleration, we have kg · m N= s2 Decimal multiples and submultiples of the units used in the International System usually involve a factor of 103 and are formed by the addition of prefixes given in Table 1.1. The unit of pressure N/m2 is also referred to as a pascal (Pa), and a mass of 1000 kg is sometimes called a tonne (or a metric ton). In the Imperial system, the basic unit of length is the foot (ft), the basic unit of force is the pound (lb), and the basic unit of time is the second (s). Mass is a derived unit. A unit mass, also known as a slug, is defined as the mass that when subjected to Table 1.1 Prefixes for multiples and submultiples of basic units in SI. Factor

Prefix

Symbol

109 106 103 10−3 10−6 10−9

giga mega kilo milli micro nano

G M k m µ n

12

Dynamics of structures

Table 1.2 Conversion between Imperial and SI units. Item

Imperial to SI

Acceleration Area

1 ft/s = 0.3048 m/s 1 ft2 = 0.0929 m2 1 in2 = 645.16 mm2 1 lb = 4.448 N 1 kip = 4.448 kN 1 ft = 0.3048 m 1 in. = 25.4 mm 1 mile = 1.609 km 1 lb = 0.4536 kg 1 lb/ft = 1.488 kg/m 1 lb/ft2 = 4.882 kg/m2 1 lb/in2 = 703.1 kg/m2 1 lb/ft3 = 16.02 kg/m3 1 lb/in3 = 27.680 Mg/m3 1 in4 = 416.230 × 103 mm4 1 in3 = 16387 mm3 1 ksi = 6.895 MPa 1 psf = 47.88 Pa 1 psi = 6.895 kPa 1 ft·kip = 1.356 kN·m 1 in3 = 16,387 mm3 1 ft3 = 28.316 × 10−3 m3

Force Length Mass Mass per unit length Mass per unit area Mass density Moment of inertia Section modulus Pressure or stress Torque or moment Volume

2

SI to Imperial 2

1 m/s2 = 3.2808 ft/s2 1 m2 = 10.764 ft2 1 mm2 = 1.55 × 10−3 in2 1 N = 0.2248 lb 1 kN = 0.2248 kip 1 m = 3.2808 ft 1 mm = 0.03937 in. 1 km = 0.6215 mile 1 kg = 2.2046 lb 1 kg/m = 0.672 lb/ft 1 kg/m2 = 0.2048 lb/ft2 1 kg/m2 = 1.422 × 10−3 lb/in2 1 kg/m3 = 0.06242 lb/ft3 1 Mg/m3 = 0.03613 lb/in3 1 mm4 = 2.4 × 10−6 in4 1 mm3 = 0.06102 × 10−3 in3 1 MPa = 0.1450 ksi 1 Pa = 0.02089 psf 1 kPa = 0.1450 psi 1 kN·m = 0.7376 ft·kip 1 mm3 = 0.06102 × 10−3 in3 1 m3 = 35.32 ft3

a force of 1 lb will be accelerated at the rate of 1 ft/s2 . Instead of specifying the mass of a body, it is usual to specify its weight, which is equal to the force of gravitation exerted on the mass. This requires the definition of a standard gravity constant, g, the acceleration due to gravity, which is taken as 32.174 ft/s2 . The value of g in SI units is 9.8066 m/s2 . It is readily seen that slug =

lb · s2 ft

Multiples or fractions of the basic Imperial units most commonly used in engineering practice are inches for length and kilopounds or kips (1000 lb) for force. Examples and exercises in this book use both SI and Imperial units. Table 1.2 will assist in conversion from one system of units to the other.

1.6

ORGANIZATION OF THE TEXT

This book is devoted to a study of the analysis of engineering structures excited by time-varying disturbances. The material is divided into five parts and twenty chapters. Part One of the book deals with the formulation of equations of motion, primarily for discrete single- and multi-degree-of-freedom systems. Part Two is devoted to the

Introduction 13

solution of the equation of motion for a single-degree-of-freedom system. Part Three deals with the solution of differential equations of motion governing the response of discrete multi-degree-of-freedom systems. Formulation of the equations of motion for continuous or distributed parameter systems is discussed in Part Four, which also describes the methods used in the solution of such equations. Part Five deals with several special topics including finite element methods, component mode synthesis, and analysis of nonlinear response. The contents of the individual chapters are described briefly in the following paragraphs. 1

2

3

4

5

6

In Chapter 1 we describe the objectives of the study of dynamic response of engineering structures and the importance of such a study in their design. The nature of dynamic forces acting on engineering structures is discussed and considerations relevant to the mathematical modeling of structures are described. Chapter 2 deals with the formulation of the equations of motion for a single-degree-of-freedom system. The system properties governing the response as well as the internal and external forces acting on a dynamic system are described. D’Alembert’s principle, which converts a dynamic problem into an equivalent problem of static equilibrium, is introduced. The governing differential equation is then derived by using either Newton’s vectorial mechanics or the principle of virtual displacement. Methods of idealizing a continuous system or a discrete multi-degree-of-freedom system by an equivalent single-degree-offreedom system are presented. Finally, the effects of gravity load, axial forces, and support motion on the governing equation are discussed. In Chapter 3 we describe the formulation of the equations of motion for a multi-degree-of-freedom system primarily through the principles of vectorial mechanics. As in the case of a single-degree-of-freedom system, the internal and external forces acting on the system are identified. Application of the Ritz method to the modeling of continuous and discrete systems is discussed. A brief description is also provided of coordinate transformation and the static condensation of stiffness matrix. In Chapter 4 we provide an introduction to the principles of analytical mechanics and their application to the formulation of equations of motion. The concepts of generalized coordinates, constraints, and work function are introduced. It is shown that the response of a dynamical system can be described through the scalar functions of work and energy, and the derivation of the Hamilton’s and Lagrange’s equations is presented. Chapter 5 deals with the analysis of free-vibration response of a single-degreeof-freedom system. Both undamped and damped systems are treated. Various types of damping mechanisms, such as viscous damping, structural damping, and Coulomb damping, are discussed. The application of the phase plane diagram to the analysis of free-vibration response of damped and undamped system is described. Chapter 6 deals with the response of a single-degree-of-freedom system to a harmonic excitation. The phenomenon of resonance is discussed. Analysis of vibration transmission from a structure to its support, and vice versa, is described. Procedures for computing the energy dissipated through damping resistance are presented. Finally, methods of measurement of damping are described.

14

7

8

9

10

11

12

13

14

15

16 17

Dynamics of structures

Response to general dynamic loading and transient response of single-degree-offreedom systems is presented in Chapter 7. In particular, response to an impulsive force and to shock loading is discussed. The concept of response spectrum is introduced and the application of response spectra to the analysis of the response to ground motion is described. In Chapter 8 we present a detailed review of the approximate and numerical methods for the analysis of single-degree-of-freedom systems. The presentation includes the Rayleigh method, numerical evaluation of the Duhamel’s integral, and direct numerical integration of the equation of motion. Errors involved in numerical integration method and the performance of various integration schemes are discussed. Chapter 9 deals with the frequency-domain analysis of single-degree-of-freedom systems. Response to a periodic load and the Fourier series representation of a periodic load are discussed. Analysis of response to a general nonperiodic load through Fourier transform method is described. Applications of discrete Fourier transform and fast Fourier transform are presented. Chapter 10 is devoted to the free-vibration response of multi-degree-of-freedom systems. The eigenvalue problem associated with free-vibration response is discussed and concepts of mode shapes and frequencies are described. Application of the mode superposition method to the solution of undamped and damped free vibrations of multi-degree-of-freedom systems is presented. In Chapter 11 we describe the various methods for the solution of eigenvalue problem of structural dynamics. The solution methods include transformation methods, iteration methods, and the determinant search method. The comparative merits of the various methods are discussed and considerations governing the selection of a method are presented. Chapter 12 deals with the forced dynamic response of multi-degree-of-freedom systems and application of the mode superposition method in the analysis of response. In Chapter 13 we present approximate and numerical methods that may be applied to the analysis of multi-degree-of-freedom systems. The methods discussed include the Rayleigh–Ritz method, direct numerical integration, and analysis in the frequency domain. In Chapter 14 we describe formulation of the equation of motion for a simple continuous system. The topics covered include flexural vibrations of a beam, axial and torsional vibrations of a rod, and lateral vibrations of a string and a shear beam. Chapter 15 deals with the analysis of free-vibration response of a simple continuous system. The associated eigenvalue problems are derived and solutions are presented for certain simple systems. Finally, the application of the mode superposition method to the analysis of free vibrations is described. Forced-vibration response of simple continuous systems using modal superposition analysis is covered in Chapter 16. In Chapter 17 we describe one-dimensional wave propagation analysis. The onedimensional wave equation is derived and the propagation of waves in a simple system, including wave reflection and refraction, is discussed. Finally, a brief presentation is given of wave propagation in a simple dispersive medium.

Introduction 15

18

19

20

Chapter 18 presents an introductory treatment of the finite element method. Discussion is limited to structural models formed by the assembly of one- and two-dimensional elements. For the two-dimensional elements, both plane strain and plane stress conditions are examined. Also presented is a discussion of the natural coordinates and their application in the derivation of the property matrices of individual elements. Chapter 19 presents a discussion of the component mode synthesis methods, also known as substructure coupling. These methods are particularly effective in the structural dynamic analysis of large complex structures, such as an aircraft, an aerospace structure, or an automobile. Chapter 20 deals with the analysis of the nonlinear response of single- and multidegree-of-freedom systems. Discussion is limited to material nonlinearity. Both explicit and implicit methods are described and illustrated by solved examples.

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Part 1

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Chapter 2

Formulation of the equations of motion: Single-degree-of-freedom systems

2.1

INTRODUCTION

The displaced configuration of many mechanical systems and structures subject to dynamic loads can be completely described by specifying the time-varying displacement along only one coordinate direction. Such systems are designated as single-degree-offreedom systems. Often the modeling of a system as a single-degree-of-freedom system is an idealization. How truly the response of the idealized model fits the true behavior depends on several factors, including the characteristics of the system, the initial conditions, the exciting force, and the response quantity of interest. Nevertheless, for a large number of systems, representation as a single-degree-of-freedom model is quite satisfactory from an engineering point of view. In view of the importance and the simplicity of single-degree-of-freedom systems, this chapter is devoted exclusively to a discussion of the formulation of equations that relate the response of such systems to one or more exciting forces. There is another equally important reason for treating the single-degree-of-freedom systems separately. As we shall see later, the analysis of the response of more complex multi-degree-offreedom systems is in many cases accomplished by obtaining the response of several related single-degree-of-freedom systems and then superimposing these responses. In this chapter the equations of motion are, in general, formulated by using the principles of the vectorial mechanics of Newton, although the principle of virtual displacement is also introduced. The procedures of Newtonian mechanics are quite adequate for simple systems. For more complex single- as well as multi-degree-offreedom systems, procedures that depend on energy principles or the principles of analytical mechanics are found to be more powerful. These procedures are discussed in detail in Chapter 4.

2.2

INERTIA FORCES

Figure 2.1 shows the simplest of all single-degree-of-freedom systems. It represents a rigid body of mass m constrained to move along the x axis in the plane of the paper. The mass is attached to a firm support by a spring of stiffness k. At any time the total time-varying force acting on the mass in a horizontal direction is denoted by Q(t). In general, it is comprised of the externally applied force p(t); the spring force fS , which depends on the displacement u of the system from a position of equilibrium;

20

Dynamics of structures

Figure 2.1 Forces on a single-degree-of-freedom system.

and the force of resistance or damping. This last force, denoted as fD , arises from air resistance and/or internal and external frictions. From Newton’s second law of motion, the applied force is equal to the rate of change of momentum d Q(t) = dt




du m dt

 (2.1)

In Equation 2.1 the momentum is expressed as the product of mass and velocity. In general, the mass of a system may also vary with time. An example of a varying mass system is a rocket in flight, where the mass of the rocket is decreasing continuously as the fuel burns out. For most mechanical systems or structures of interest to us, however, mass does not vary with time and therefore can be taken out of the differentiation. Using overdots to represent differentiation with respect to time, Equation 2.1 can be rewritten as Q(t) = mu¨ or Q(t) − mu¨ = 0

(2.2)

Quantity mu¨ has the units of a force. If we define an inertia force as having a magnitude equal to the product of mass and acceleration and a direction opposite to the direction of acceleration, we can view Equation 2.2 as an equation of equilibrium among the forces acting on a body. This principle, known as d’Alembert’s principle, converts the problem of dynamic response to an equivalent static problem involving equilibrium of forces and permits us to use for its solution all those procedures that we use for solving problems of the latter class. On a cursory glance, d’Alembert’s principle appears simply as a mathematical artifact. Its physical significance can, however, be appreciated by considering the following simple example. Figure 2.2 shows a spring balance bolted to the floor of an elevator. A body having mass m is placed on the scale. First let the elevator be at rest. The reading on the scale will indicate the weight of the body, mg, where g is the acceleration due to gravity. Now let the body be pulled upward with a force F which is less than the weight mg. The scale will record a new reading equal to mg − F. Obviously, the downward force of gravity exerted by the body is being counteracted by an upward force F. Next let the force F be removed, but let the elevator move downward with an acceleration a. The scale reading will change from mg to mg − ma. To an observer inside the elevator,

Formulation of the equations of motion: Single-degree-of-freedom systems 21

Figure 2.2 Inertia force on a moving mass.

the effect on the scale reading of a downward acceleration is no different from that of the upward force F. The quantity ma thus manifests itself as a virtual force acting in a direction opposite to the direction of acceleration. 2.3

RESULTANTS OF INERTIA FORCES ON A RIGID BODY

Structures or mechanical systems are sometimes modeled as rigid bodies connected to each other and to supports, often through deformable springs. We will find it expedient to replace the inertial forces acting on a rigid body by a set of resultant forces and moments. As an example, consider two point masses m1 and m2 connected together by a massless rigid rod as shown in Figure 2.3. The motion of this system in a plane can be described by specifying the translations of the center of mass in two mutually perpendicular directions and a rotation about that center. It is readily seen that the total inertial force due to translation in the x direction is equal to (m1 + m2 )u¨ x and acts through the center of mass. Similarly, for a y translation the total inertial force is (m1 + m2 )u¨ y . For rotation about the center of mass the two inertia forces m1 aθ¨ and m2 bθ¨ are equal and opposite and form a couple whose magnitude is (m1 a2 + m2 b2 )θ¨ . Next, consider the uniform rigid bar shown in Figure 2.4a. The bar has a mass m ¯ per unit length. Consider a motion of the bar in the y direction. All particles of the bar have an acceleration u¨ y and the inertial force on an incremental length dx is equal to m ¯ u¨ y dx. Since all such infinitesimal forces are parallel and act in the negative y direction, they can be replaced by a single resultant force fI given by  fI =

L

m ¯ u¨ y dx = mL ¯ u¨ y

0

where L is the length of the bar.

(2.3)

22

Dynamics of structures

Figure 2.3 Resultant of inertia forces on a rigid body.

Figure 2.4 Inertia forces on a uniform rigid bar: (a) translation in y direction; (b) rotation about the center of mass.

The distance of the point of action of this force from the left end of the rod is obtained by taking moments of the inertial forces  L L2 fI x¯ = m ¯ u¨ y x dx = m ¯ u¨ y 2 0 ¯ u¨ y mL ¯ u¨ y x¯ = m

L2 2

(2.4)

L 2 Thus the resultant force acts through the center of mass of the bar, which in this case coincides with the midpoint of the bar. Next consider a rotation of the bar about its center of mass. As shown in Figure 2.4b, the inertial force acting on an infinitesimal section at a distance x from the center of mass is, in this case, equal to mx ¯ θ¨ dx. The sum of all such forces is given by  L/2 ¨ fI = θ mx ¯ dx (2.5) x¯ =

−L/2

Formulation of the equations of motion: Single-degree-of-freedom systems 23

Figure 2.5 Inertia forces on a rigid bar rotating about its end.

But since x is measured from the center of mass, the integral in Equation 2.5 is zero and fI vanishes. The infinitesimal inertial force also contributes a moment about the center of mass whose value is mx ¯ 2 θ¨ dx. The total moment is given by MI = θ¨



L/2

−L/2

mx ¯ 2 dx = I0 θ¨

(2.6)

where I0 , termed the mass moment of inertia, is equal to the value of the integral, which works out to mL ¯ 3 /12 = mL2 /12, m being the total mass of the rigid bar. A comparison of Equations 2.2 and 2.6 shows that in rotational motion, the mass moment of inertia plays the same role as does mass in translational motion. The force in the latter is replaced by a moment in the former and the translational acceleration u¨ is replaced by the angular acceleration θ¨ . The mass moment of inertia is often expressed as I0 = mr2 , in which r is termed the radius of √ gyration. For a uniform rod rotating about its centroid, the radius of gyration is L/ 12. As another example of resultant inertial forces on a rigid body, consider a uniform bar rotating about its left-hand end. This motion is equivalent to a translation uy = Lθ/2 and a rotation θ both measured at the center of mass. The motions at the center of mass give rise to an inertial force mLθ¨ /2 and a moment mL2 θ¨ /12 as indicated in Figure 2.5. The force and the moment about the center of mass are equivalent to a force of mLθ¨ /2 and a moment of mL2 θ¨ /12 + L/2 · mLθ¨ /2 = mL2 θ¨ /3 at the left-hand end of the bar. The resultant inertia forces for several different rigid-body shapes are given in Figure 2.6.

24

Dynamics of structures

Figure 2.6 Resultant inertia forces in rigid-body shapes: (a) uniform bar, mass m; (b) uniform rectangular plate, mass m; (c) uniform circular plate, mass m (d) Resultant inertia forces in rigid-body shapes: (d) uniform elliptical plate, mass m.

Formulation of the equations of motion: Single-degree-of-freedom systems 25

Figure 2.6 Cotinued.

Example 2.1 Determine the mass moment of inertia of a uniform rectangular rigid plate, having a mass m ¯ per unit area, for rotation about (a) an axis perpendicular to its plane and passing through its centroid, and (b) the x axis.

Solution (a)

Referring to Figure E2.1a, the x and y components of the inertial force acting on an infinitesimal area dx dy of the plate are my ¯ θ¨ dx dy and mx ¯ θ¨ dx dy, respectively. The resultant force in the x direction is given by  Fx =

b/2



−b/2

a/2 −a/2

my ¯ θ¨ dx dy = m ¯ θ¨



b/2



a/2

y dx dy −b/2

(a)

−a/2

Because y is measured from the centroid, the integral on the right-hand side of Equation a is zero and Fx vanishes. For a similar reason, Fy is zero. A moment about the centroid however exists and is given by M = I0 θ¨ =



+b/2



−b/2

+a/2

−a/2

m ¯ θ¨ (x2 + y2 ) dx dy

a2 + b 2 θ¨ 12 a 2 + b2 θ¨ =m 12 = mab ¯

where m is the total mass of the plate. Equation b gives I0 = m(a2 + b2 )/12.

(b)

26 (b)

Dynamics of structures For rotation about the x axis, divide the plate into rigid bars of length b and width dx 2 as shown in Figure E2.1b. The inertial moment contributed by each bar is mb(b ¯ /12) dx. The total moment is given by I0x θ¨ = θ¨ =



a/2

−a/2

mb ¯

b2 dx 12

mb2 θ¨ 12

giving I0x = mb2 /12.

Figure E2.1 Inertia forces on a rigid rectangular plate.

2.4

SPRING FORCES

An elastic body undergoing deformation under the action of external forces sets up internal forces that resist the deformation. These forces of elastic constraints, alternatively known as spring forces, are present irrespective of whether the deformation is a result of static or dynamic forces acting on the body. The simplest example of the force of elastic constraint is provided by a helical spring shown in Figure 2.7. When stretched by a force applied at its end, the spring resists the deformation by an internal force which is related to the magnitude of the deformation. The relationship between the spring force and the displacement of the spring may take the form shown in Figure 2.7. In general, this relationship is nonlinear. However, for many systems, particularly when the deformations are small, the force– displacement relationship can be idealized by a straight line. The slope of this straight line is denoted by k and is called the spring constant or the spring rate. Thus, for a linear spring, the spring constant can be defined as the force required to cause a unit displacement and the total spring force is given by fS = ku

(2.7)

Formulation of the equations of motion: Single-degree-of-freedom systems 27

Figure 2.7 Force-displacement relationship for a helical spring.

Figure 2.8 Definition of spring forces in elastic structures: (a) cantilever beam; (b) portal frame.

A similar definition of spring force can be applied to the forces of elastic constraints exerted by bodies of other form. For example, consider the cantilever beam shown in Figure 2.8a, and let the coordinate of interest be along a vertical at the end of the beam. A vertical load P applied at the end of the beam will cause a displacement in the direction of the load of PL3 /3EI, where L is the length of the beam, E the modulus of elasticity, and I the moment of inertia of the beam cross section. The spring constant, the force required to cause a unit displacement, is therefore 3EI/L3 . Figure 2.8b shows a portal frame consisting of a rigid beam supported by two similar uniform columns each of length L, both fixed at their base. For a coordinate in the horizontal direction at the level of the beam, the spring constant is easily determined as 24EI/L3 . As stated earlier, for large deformations the relationship between the spring force and the displacement of the spring becomes nonlinear. For most structural materials this leads to a softening of the spring, or a reduction in the slope of the force-displacement diagram. In addition, the path followed by the force-displacement diagram during unloading is different from that followed during loading. As an illustration consider the moment-rotation curves shown in Figure 2.9. These curves were obtained during an experiment on a steel-concrete composite beam subjected to cyclic loading. The moment-rotation relationship is quite complex, the rotation corresponding to a given moment is not unique but depends on the loading history. During

28

Dynamics of structures

Figure 2.9 Moment-rotation relationship for a steel-concrete composite beam. (From J. L. Humar, “Composite Beams under Cyclic Loading,’’ Journal of Structural Division, ASCE, Vol. 105, 1979, pp. 1949–1965.)

each cycle of loading the moment-rotation relationship forms a closed loop. The area enclosed by this loop represents the energy lost during that cycle of loading. This energy loss is caused by plastic deformation of the material. It is evident that the dynamic analysis of a system in which the structural material is strained into the inelastic range would be quite complex. Nevertheless analysis in the nonlinear range is often of practical interest. For example, under earthquake loading structures are usually expected to be strained beyond the elastic limit and the energy loss in plastic deformation must be considered in design. In Chapter 20 we briefly discuss the analysis of dynamic response of nonlinear structures. Example 2.2 Find the equivalent stiffness of a system comprised of two linear springs (Fig. E2.2) having spring constants k1 and k2 when the two springs are connected (a) in parallel, and (b) in series.

Solution (a)

Assume that the system is configured so that the springs undergo equal displacement. For a displacement u, F1 = k1 u and F2 = k2 u. Therefore, F = (k1 + k2 )u and the spring constant is given by k=

F = k1 + k2 u

Formulation of the equations of motion: Single-degree-of-freedom systems 29

Figure E2.2 Definition of equivalent stiffness: (a) springs connected in parallel; (b) springs connected in series.

(b)

In this case the force in each spring should be the same and equal to the externally applied force F. The extension of the first spring is 1 = F/k1 . The extension of the second spring is 2 = F/k2 . Thus
 = 1 + 2 = F

1 1 + k1 k2



and k=

2.5

F 1 k 1 k2 = =  1/k1 + 1/k2 k1 + k 2

DAMPING FORCES

As stated earlier, the motion of a body is in practice resisted by several kinds of damping forces. These forces are always opposed to the direction of motion, but their characteristics are difficult to define or to measure. Usually, the magnitude of a damping force is small in comparison to the force of inertia and the spring force. Despite this, damping force may significantly affect the response. Also, in many mechanical systems, damping devices are incorporated on purpose. These devices help in controlling the vibrations of the system and are usually designed to provide a substantial amount of damping. For example, the shock absorbers in an automobile are devices which by providing large damping forces, cut down the unwanted vibrations. A damping force may result from the resistance offered by air. When the velocity is small, the resistance offered by a fluid or a gas is proportional to the velocity. A resisting force of this nature is called viscous damping force and is given by fD = cu˙

(2.8)

where c is a constant of proportionality called the damping coefficient. The damping force is, of course, always opposed to the direction of motion. When the velocity is large, the damping force due to air resistance becomes proportional to the square of the velocity. Therefore, a viscous damping force is either a linear or a nonlinear function of the velocity. As we will see later, the velocity of

30

Dynamics of structures

a vibrating system is, in general, proportional to its frequency, and hence a viscous damping force increases with the frequency of vibration. Forces resisting a motion may also arise from dry friction along a nonlubricated surface. A resisting force of this nature is called the force of Coulomb friction. It is usually assumed to be a force of constant magnitude but opposed to the direction of motion. In addition to the forces of air resistance and external friction, damping forces also arise because of imperfect elasticity or internal friction within the body, even when the stresses in the material do not exceed its elastic limit. Observations have shown that the magnitude of such a force is independent of the frequency but is proportional to the amplitude of vibration or to the displacement. Resisting forces arising from internal friction are called forces of hysteretic or structural damping. In real structures a number of different sources of energy dissipation exist. Energy may be lost due to repeated movements along internal cracks, such as those existing in reinforced concrete and masonry structures. Energy is dissipated through friction when slip takes place in the joints of steel structures. Relative movements at the interfaces between nonstructural elements, such as in-fill walls, and the surrounding structural elements also cause loss of energy. The foregoing discussion indicates that damping forces are complex in nature and difficult to determine. In general, viscous damping forces are easiest to handle mathematically and are known to provide analytical results for the response of a system that conform reasonably well to experimental observations. It is therefore usual to model the forces of resistance as being caused by viscous damping. The viscous damping mechanism is indicated by a dashpot, as shown in Figure 2.10. The general nature of damping force is represented by a nonlinear relationship between the force and the velocity. The relationship can, however, be approximated by a straight line in which case the damping force is given by Equation 2.8. As noted in Section 2.4 energy loss may also occur through repeated cyclic loading of structural elements in the inelastic range. One may account for such energy loss by defining a viscous damping model with an appropriately selected value of the damping coefficient c. However, such a model is unable to correctly represent the mechanism of energy loss. Energy loss through plastic flow in repeated cyclic loading is best accounted for by adopting an appropriate model to represent the nonlinear displacement versus spring force relationship. The damping mechanism used in the mathematical model then accounts for the loss of energy through sources other than inelastic deformations.

Figure 2.10 Representation of a viscous damping force.

Formulation of the equations of motion: Single-degree-of-freedom systems 31

2.6

PRINCIPLE OF VIRTUAL DISPLACEMENT

As stated earlier, once the inertial forces have been identified and introduced according to d’Alembert’s principle, a dynamic problem can be treated as a problem of static equilibrium, and the equations of motion can be obtained by direct formulation of the equations of equilibrium. The latter equations are obtained by the well-known methods of vectorial mechanics. For complex systems, however, use of these methods may not be straightforward. In such cases, the application of the principle of virtual displacement may greatly simplify the problem. The principle of virtual displacement, in fact, belongs to the field of analytical mechanics,which deals with the scalar quantities of work and energy. It is, however, of sufficient simplicity for us to discuss here, in advance of a general and more detailed discussion of the methods of analytical mechanics in Chapter 4. Suppose that a mechanical or structural system is in equilibrium under a set of externally applied forces and the forces of constraints, such as, for example, the reactions at the supports. If the system is subjected to virtual or imaginary displacements that are compatible with the constraints in the system, the applied forces, as well as the internal forces,will do work in riding through the displacements. We denote the work done by the external forces as δWe and that done by the internal forces as δWi . The symbol δW is used rather than W to signify that the work done is imaginary or virtual and not real. Let the displacements be applied gradually so that no kinetic energies are developed. Also assume that no heat is generated in the system and that the process is adiabatic, so that no heat is added or withdrawn. Under these conditions the law of conservation of energy holds, giving δWe + δWi = 0.0

(2.9)

The definition of internal work needs careful consideration. It can be illustrated by a simple example of a particle attached to a linear spring of constant k, shown in Figure 2.11. The system is in equilibrium under an external force F. Now let the particle undergo a virtual displacement δu as shown in the figure. The virtual work done by the external force is F δu. Internal work is done by the spring force which results from the deformation, u, of the spring. This force is equal to −ku, and acts

Figure 2.11 Definition of virtual work.

32

Dynamics of structures

on the particle. The virtual work done by the spring force is equal to −ku δu and the virtual work equation takes the form (F − ku)δu = 0.0

(2.10)

Alternatively, a force equal to ku can be assumed to be acting on the end of the spring. This force is, of course, equal and opposite to the force ku acting on the particle. The virtual work done by this force is ku δu. A work of this nature is sometimes referred to as the work done by forces acting on the internal elements. It is denoted by δWei and is negative of the internal work. Using this alternative definition, the virtual work equation can be stated as δWe = δWei

(2.11)

When the system under consideration is rigid, the internal deformations and hence δWei are zero, and the virtual work equation takes the simpler form δWe = 0.0

(2.12)

The principle of virtual displacement can now be stated as follows. If a deformable system in equilibrium under a set of forces is given a virtual displacement that is compatible with the constraints in the system, the sum of the total external virtual work and the internal virtual work is zero. The principle is of sufficient generality and applies equally well to linear or nonlinear elastic bodies. The only restriction is that the displacements be compatible with the constraints in the system. In calculating the internal virtual work in deformable bodies, it simplifies the problem if the internal forces can be assumed to remain constant as they ride through the virtual displacements. In general, this assumption is true only if the virtual displacements are small, so that the geometry of the system is not materially altered. It should be noted, however, that virtual displacements do not really exist; their imposition is simply a mathematical experiment that helps us determine the equations of equilibrium. Since the magnitude of a virtual displacement is arbitrary, it is only logical to assume that it is small. The virtual work equations obtained by imposing appropriate virtual displacements on the system lead to the desired equations of equilibrium. In the following examples, we illustrate the application of the principle of virtual displacement to the formulation of such equations. Example 2.3 Determine the reaction at A in the simply supported beam shown in Figure E2.3.

Solution A virtual displacement that is compatible with the constraints in the system will involve no vertical displacement at support A. The reaction R at support will not thus appear in any virtual work equation and cannot therefore be determined by the principle of virtual displacement. To be able to determine the reaction at A, we release the displacement constraint at A and replace it by the unknown force R. There is now only one constraint in the released structure, that is, a zero displacement at support B. We select a virtual displacement pattern caused by a rigid-body rotation of the beam about B. This displacement is compatible with the constraint in the system.

Formulation of the equations of motion: Single-degree-of-freedom systems 33

Figure E2.3 (a) Simply supported beam; (b) virtual displacements in the released beam.

Further, it causes no internal deformations so that δWi is zero and the virtual work equation becomes δWe = 0.0 or RLδ θ − Paδ θ = 0.0 which gives R=

Pa L

Example 2.4 For the frame shown in Figure E2.4a, determine the spring force due to the applied loads W and H.

Solution To solve the problem, we first replace the spring by a pair of equal and opposite forces X and then give the resulting system a small virtual displacement δφ. The virtual displacements of the entire system are shown in Figure E2.4b, and we note that they are compatible with the constraints in the system. The height of the load W above the base AC is given by h = (a + b) sin φ

(a)

The change in h as a result of the change in φ is given by δh = (a + b) cos φδφ

(b)

34

Dynamics of structures

Figure E2.4 Formulation of the equations of equilibrium by the principle of virtual displacement: (a) A-frame in equilibrium; (b) virtual displacements; (c) free body diagrams.

In a similar manner, the horizontal distance of B from A is (a + b)cos φ, and changes by −(a + b)sin φ δφ. The horizontal distance to D from a vertical axis at A is a cos φ and this changes by −a sin φ δφ as φ increases by δφ. The horizontal distance to E from the vertical axis at A is (a + 2b)cos φ. This changes by −(a + 2b)sin φ δφ. Since we have replaced the spring by a pair of forces, the resulting system consists of rigid bodies and the simpler form of the virtual work equation (Eq. 2.12) can be used, giving −W(a + b)cos φ δφ + X(−a sin φ δφ) + X(a + 2b)sin φ δφ − H(a + b)sin φ δφ = 0

(c)

Equation c gives X=

W(a + b) a+b cot φ + H (2b) 2b

(d)

Formulation of the equations of motion: Single-degree-of-freedom systems 35 It is instructive to solve the same problem by a direct formulation of the equation of equilibrium. By equating the moment of all the forces about A to zero, the vertical support reaction at C is obtained as W/2 + Hsin φ/(2cos φ). The vertical and the horizontal reactions at A are W/2 − {Htan φ}/2 and H, respectively. Free-body diagrams can now be drawn for rods AB and BC and are shown in Figure E2.4c. The horizontal reaction at the hinge at B is obtained by considering the equilibrium of rod BC and taking moments about E.
 W H a + b cos φ R= + tan φ 2 2 b sin φ =

a+b W (a + b)cot φ + H 2b 2b

(e)

The spring force X is equal to R. Both the virtual displacement and the direct equilibrium solutions to the problem are based on the assumption that the initial angle φ which the leg AB makes with the horizontal does not change appreciably with the application of the two forces. In this simple example the advantage of using the method of virtual displacement is not at once evident. We may note, however, that in direct formulation of the equation of equilibrium, we had to determine the support reactions, even though they were of no interest to us. On the other hand, these forces did not appear in our virtual work equations, because the compatible virtual displacements in the directions of the support reactions were zero. The possibility of avoiding the determination of the forces of constraints can simplify the problem significantly when the system being analyzed is complex.

2.7

FORMULATION OF THE EQUATIONS OF MOTION

Having discussed the characteristics of the forces acting in a dynamic system, we are now in a position to formulate the equations of motion. Following d’Alembert’s principle, the dynamic problem is first converted to a problem of the equilibrium of forces by introducing appropriate inertia forces. The equations of dynamic equilibrium are then obtained either by the direct methods of vectorial mechanics or by the application of the principle of virtual displacements. In discussing the formulation of equation of motion, it is convenient to classify the system into one of the following four categories 1 2 3 4

Systems with localized mass and localized stiffness Systems with localized mass but distributed stiffness Systems with distributed mass but localized stiffness Systems with distributed mass and distributed stiffness

The formulation of the equation of motion for each of the foregoing categories is discussed in the following paragraphs.

2.7.1 Systems with localized mass and localized stiffness Figure 2.12a shows the simplest single-degree-of-freedom system. The mechanism providing the force of elastic constraint is localized in the massless spring. The mass can

36

Dynamics of structures

Figure 2.12 Dynamic equilibrium of a simple single-degree-of-freedom system.

be assumed to be concentrated at the mass center of the rigid block. The mechanism providing the force of resistance is represented by a dashpot. The spring force, the damping force, the force of inertia, and the externally applied force are all identified on the diagram. The equation of motion is obtained directly on equating the sum of the forces acting along the x axis to zero. Thus fI + fD + fS = p(t)

(2.13)

As an alternative, we may reason that the spring forces fS must be balanced by an equal and opposite force, p1 = fS , as shown in Figure 2.12b. In a similar manner, a force p2 = fD is required to counter the force of damping, and force p3 = fI is required to balance the inertia force. The total external force must equal p1 + p2 + p3 , leading to Equation 2.9. We will find that this line of reasoning makes it simpler to visualize the formulation of the equations of motion for multi-degree-of-freedom systems. If the spring is assumed to be linear with spring constants k and the damping is viscous with a damping coefficient c, Equation 2.13 can be written as mu¨ + cu˙ + ku = p(t)

(2.14)

Formulation of the equations of motion: Single-degree-of-freedom systems 37

Figure 2.13 Vibration of a point mass attached to a light cantilever beam.

Equation 2.14 is linear because coefficients m, k, and c are constant and because the power of u and its time derivatives is no more than 1. Further, the equation is a second order differential equation. Its solution will give u as a function of time.

2.7.2 Systems with localized mass but distributed stiffness A simple system in which the stiffness is distributed but the mass is localized is shown in Figure 2.13. It consist of a light cantilever beam of flexural rigidity EI, to the end of which is attached a point mass m. The beam is assumed to be axially rigid and is constrained to deform in the plane of the paper. The system shown in Figure 2.13 is a single-degree-of-freedom system, and as in the case of a system with localized mass as well as localized stiffness, we define the single degree of freedom along the coordinate direction in which the mass is free to move. In the present case this direction is along a vertical through the tip of the beam. The inertia force in the system acts in the coordinated direction we have chosen and is proportional to the vertical acceleration of the mass. Further, we assume that the external force also acts in the same direction. Once the displacement along the selected coordinate direction is known the displaced shape of the beam can be determined by standard methods of structural analysis. These methods also permit us to obtain the force of elastic restraint acting along the selected coordinate direction. This is the case, for example, for the cantilever beam and the portal frame shown in Figure 2.8. The foregoing discussion shows that the system of Figure 2.13a is entirely equivalent to the spring–mass system shown in Figure 2.13b. The equation of motion is obtained directly by equating the sum of vertical forces to zero mu¨ +

3EI u = p(t) L3

(2.15)

Again, the force of gravity has been ignored in the formulation. This is admissible provided it is understood that u is measured from the position of static equilibrium. Example 2.5 The torque pendulum shown in Figure E2.5 consists of a light shaft of uniform circular cross section having diameter d and length L. The material used to build the shaft has a modulus of rigidity G. A heavy flywheel having a mass m and a radius of gyration r is attached to the end of the shaft. Find the equation of free vibrations of the torque pendulum.

38

Dynamics of structures

Figure E2.5 Vibrations of flywheel.

Solution We select the angle of twist at the end of the shaft, θ, as the single degree of freedom of the system. A torque T applied at the end of the shaft will twist it by an angle θ = TL/GJ in which J, the polar moment of inertia of the shaft, is equal to πd 4 /32. The torque required to cause a twist θ can therefore be expressed as T = GJθ/L. The inertial moment induced by an angular acceleration θ¨ of the flywheel is given by mr2 θ¨ . When there is no externally applied force, the equation of equilibrium for angular motion is given by mr2 θ¨ +

GJ θ=0 L

2.7.3 Systems with distributed mass but localized stiffness An example of such a system is shown in Figure 2.14. The system consists of a uniform rigid bar of mass m hinged at its left-hand end and suspended by a spring at the right end. The force due to elastic constraint is localized in a spring. However, the mass is uniformly distributed along the rigid bar. As a result, the inertia forces are also distributed along the length. We may be able to replace the distributed inertia forces by their resultants using the methods described in Section 2.3. However, in most cases we are still left with more than one resultant force. In spite of the fact that the inertia forces are distributed, the system of Figure 2.14 has only one degree of freedom, because we can completely define the displaced shape of the system in terms of the displacement along a single coordinate. Thus if z(t) is the unknown displacement along the selected coordinate and ψ(x) is a selected shape function, the displacement of the system is given by u(x, t) = z(t)ψ(x)

(2.16)

For the bar shown in Figure 2.14 we may select the rotation at the hinge, or the vertical displacement at the tip of the rod, as the degree of freedom of the system. Let us choose the latter and denote the displacement along the selected coordinate direction

Formulation of the equations of motion: Single-degree-of-freedom systems 39

Figure 2.14 Vibration of a rigid bar.

as z(t); then because the given bar is rigid, the displacement shape function is readily seen to be x (2.17) ψ(x) = L where x is the distance from the hinge. Once we solve for the unknown displacement z(t), the displacement of the bar is fully determined. A coordinate such as z(t) is known as a generalized coordinate and the associated system model as a generalized single-degree-of-freedom system. The displaced shape of the system is determined in terms of the generalized coordinate through the selected shape function. When the system consists of rigid body assemblages, the exact shape function is readily found. Continuing with the example of rigid bar, it is readily seen that the bar has an angular acceleration of z¨ /L and a vertical translational acceleration of z¨ /2 at its mass center. The corresponding forces of inertia are obtained from Section 2.3 and are indicated on the diagram. Taking moments about the hinge, we obtain I0 θ¨ + m or




z¨ L b2 + c˙z + kzL = p(t)a 2 2 L

(2.18a)


2
 mL mL cb kL + z¨ + z˙ + z = p(t) 12a 4a La a

m∗ z¨ + c∗ z˙ + k∗ z = p∗ (t)

(2.18b)

40

Dynamics of structures

where θ¨ = z¨ /L is the angular acceleration, m∗ = mL/3a is known as the generalized mass, c∗ = cb2 /La is the generalized damping constant, k∗ = kL/a is the generalized spring constant, and p∗ = p is the generalized force. It should be noted that in addition to the forces indicated in Figure 2.14b, the force of gravity equal to mg acts downward through the center of mass of the rigid rod. This force and the spring force set up to balance it can both be ignored in the formulation of the equation of motion as long as it is assumed that the displacement u is measured from the position of equilibrium of the rod when acted upon only by the force of gravity. This point is dealt with in some detail in a later section of this chapter. Example 2.6 The swinging frame of a balancing machine is shown diagrammatically in Figure E2.6a. Lever ABC, which has a nonuniform section, has a mass of 5.5 kg and radius of gyration 75 mm about its mass center, which is located at B. Lever DF is of uniform section and has a mass of 1 kg. A mass of 0.7 kg is attached to lever DF at point E. The mass of the connecting link CF may be ignored. The springs at A and C both have a stiffness of 5.6 kN/m. Obtain the equation of motion for small vibrations of the frame about a position of equilibrium.

Solution The displaced shape of the frame is completely determined by specifying the displacement along one coordinate direction. Therefore, the frame has a single degree of freedom. We select the vertical deflection at F as the single coordinate. The equation of motion will be obtained in terms of the selected coordinate, which we shall denote by z. Also, in our discussion, we shall use the basic units of measurement: meters for length, kilograms for mass, and newtons for force. The displaced shape of the frame for a vertical deflection z at F is shown in Figure E2.6b along with all the forces acting on the frame. Corresponding to a vertical upward displacement u meters at C, the spring attached to that point exerts a downward force of 5600u newtons. Point A moves a distance z/6 meter downward. The spring at A resists this movement by an upward force of 5600 × u/6 = 933.3u N. The inertial moment opposing the rotation of bar ABC acts at B and is given by (MI )B = IB θ¨ = 5.5 × 0.0752 ×

z¨ 0.3

= 0.1031¨z

where θ is the rotation of bar ABC about B. The inertial moment at D due to the rotation of bar DEF is obtained from (MI )D = ID φ¨ z¨ 1 × 0.452 × 3 0.45 = 0.15¨z

=

Formulation of the equations of motion: Single-degree-of-freedom systems 41

Figure E2.6 (a) Swinging frame; (b) displacements and forces in the vibrating frame. where φ represents the rotation of bar DEF about D. Finally, the downward force of inertia at E due to the motion of mass attached at that point is given by PI = mu¨ E 0.35 z¨ 0.45 = 0.5444¨z = 0.7 ×

To obtain the equation of motion, we impart a virtual displacement that is compatible with the constraints on the frame and equate the resulting virtual work to zero. A small vertical

42

Dynamics of structures

deflection δz at point F is an appropriate virtual deflection for our purpose. The displacements produced in the framedue to the deflection δz are identical to those shown in Figure E2.5b with z replaced by δz. The virtual work equation becomes 0.1031¨z ×

δz δz 0.35 δz + 0.15¨z × + 0.5444¨z δz × + 933.3z × + 5600z δz = 0 0.3 0.45 0.45 6

On canceling out δz from the virtual work equation, we get the following equation of motion 1.1004¨z + 5755.6z = 0 It is worth noting that we did not have to account for the reaction forces at supports B and D or for the force in link CF. These forces do not, in fact, appear in the virtual work expression and are not of interest to us.

2.7.4 Systems with distributed stiffness and distributed mass Consider the horizontal bar of Figure 2.15. The bar has a mass m ¯ per unit length, an axial rigidity EA, and is vibrating in the axial direction under the action of a tip force P(t). The inertia forces are now distributed along the length of the bar. Because the flexibility is also distributed along the length, one needs to know the exact distribution of the inertia forces to be able to determine the displaced shape of the bar. One can imagine the bar as being composed of a large number of sections each of a very small

Figure 2.15 Axial vibration of a bar.

Formulation of the equations of motion: Single-degree-of-freedom systems 43

length x = L/N, where N is the total number of sections. The inertia force on the ith section is equal to m ¯ x u¨ i . To obtain these inertia forces, one must determine the acceleration u¨ i for all values of i from 1 to N. The number of unknown displacements or accelerations that must be determined is equal to the number of degrees of freedom. Therefore, the system has as many degrees of freedom as there are sections in it. Theoretically, the bar has an infinite number of degrees of freedom. We can, however, obtain an approximate picture of the behavior of this vibrating bar by dividing it into a finite number of sections and then assuming that displacements of all points within a section are the same and are equal to the displacement of the center of the section. This is equivalent to lumping the mass of a section at its midpoint. We now have a model whose behavior will approximate the true behavior of the bar. This model, called a lumped mass model, is shown in Figure 2.15b. Figure 2.15c shows an alternative method of lumping the masses. The number of displacement coordinates that we must know in order to analyze a lumped mass model, and hence the number of degrees of freedom the model has, is equal to the number of mass points. To improve the accuracy of the model, we must increase the number of subdivisions. This will, however, increase the number of degrees of freedom and hence the complexity of the analysis. As an alternative to the lumped mass model, we can describe the motion of the system by assuming that the displacement of the rod u is given by u(x, t) = z(t)ψ(x)

(2.19)

where the z(t) is a function of time and ψ(x) a function of the spatial coordinate x. If we make an appropriate choice for the shape function ψ(x), the only unknown is z(t), and the system reduces to a single-degree-of-freedom system. As in the case of a rigid-body system with distributed mass, z is referred to as a generalized coordinate. Unlike the case of a rigid-body assemblage the exact shape function is not determined and judgment must be used in selecting the shape function. It may be noted that z is not necessarily a coordinate which we can physically measure. Nevertheless, once z is determined, Equation 2.19 leads us to u. The concept of the generalized coordinates is explored more fully in Chapter 4. Continuing with our example of the axial vibration of a rod, let us assume that a shape function ψ(x) has been selected and that the displacement of the rod is therefore given by Equation 2.19. The inertia force acting on a small section of the rod is now given by fI = m ¯ dx u(x, ¨ t) = m ¯ dx z¨ (t)ψ(x)

(2.20)

and the spring force by fS = EA(x)

∂u dψ = EA(x)z(t) ∂x dx

(2.21)

We now give an admissible virtual displacement to the rod. Note that such an admissible displacement function should lead to small displacements; as well, it should satisfy the constraint on the system, that is, it should give a zero displacement at the

44

Dynamics of structures

fixed end. Now suppose that ψ(x) was selected to satisfy a similar constraint; then δz ψ(x) is clearly an admissible virtual displacement. The elongation of the infinitesimal section associated with the virtual displacement δzψ(x) is δz

dψ dx dx

(2.22)

The virtual work done by the elastic spring forces acting on the section is given by d(δWei ) = fS δz

dψ dx dx

(2.23)

Substituting for fS from Equation 2.21 and integrating over the length, we obtain the total virtual work done on the internal elements 

L

δWei = δz z(t)

EA(x){ψ (x)} dx 2

(2.24)

0

where a prime denotes differentiation with respect to x. The external forces are comprised of body forces due to inertia (Eq. 2.20) and the tip force P(t). The virtual work done by these forces is given by  δWe = −δz¨z(t)

L

2 m(x){ψ(x)} ¯ dx + P(t) δz ψ(L)

(2.25)

0

In accordance with the principle of virtual displacement (Eq. 2.11) the total external virtual work must be equal to the virtual work done on the internal elements. On equating Equations 2.24 and 2.25 and canceling out δz, we get 

L

z¨ (t)



L

2

m(x){ψ(x)} ¯ dx + z(t)

0

EA(x){ψ (x)} dx = P(t)ψ(L) 2

(2.26)

0

or m∗ z¨ (t) + k∗ z(t) = p∗

(2.27)

where ∗



L

m = k∗ =



2 m(x){ψ(x)} ¯ dx

0 L

EA(x){ψ (x)} dx

0

and p∗ = P(t)ψ(L)

2

Formulation of the equations of motion: Single-degree-of-freedom systems 45

Mass m∗ , which is associated with the generalized coordinate z, is the generalized mass; similarly, k∗ is the generalized stiffness and p∗ is the generalized force. Example 2.7 Write the equation of motion for the free axial vibration of the uniform bar shown in Figure E2.7a using a displacement shape function given by (a) ψ(x) = x/L, and (b) ψ(x) = sin(πx/2L).

Solution (a)

The generalized mass is given by

m∗ =



L

 x 2

m ¯

0

L

dx =

1 m 3

(a)

where m is the total mass of the bar. The generalized stiffness k∗ is given by

k∗ =




2 1 EA dx = L L

L

EA 0

(b)

Figure E2.7 Axial vibration of a uniform bar: (a) uniform bar; (b) shape function ψ = x/L; (c) shape function ψ = sin πx/2L.

46

Dynamics of structures The equation of motion is 1 EA m¨z(t) + z(t) = 0 3 L

(b)

(c)

The forces acting on a small section of the bar are indicated in Figure E2.7b. It is noted that regardless of the value of z, the forces are not in equilibrium. In general, with the procedure being used, equilibrium will be satisfied only in an average sense and the results obtained are therefore necessarily approximate. The generalized mass is in this case given by m∗ =



L

0

 m πx 2 L m ¯ sin dx = m ¯ = 2L 2 2

(d)

The generalized stiffness k∗ is obtained from k∗ =



 π πx 2 π2 EA cos dx = 2L 2L 8L

L

EA 0

(e)

The equation of motion is m¨z(t) +

π2 EA z(t) = 0 4 L

(f)

By substitution it can  be seen that z(t) = Gsin(ωt + φ) is a solution of Equation f provided that ω = (π/2) EA/mL. Parameters G and φ are as yet undetermined constants. The displaced shape of the bar is given by u(x, t) = Gsin(ωt + φ) sin

πx 2L

(g)

The inertia force acting on a section of the bar is now given by πx ¨ m fI = z(t) dx ¯ sin 2L πx dx ¯ sin = −ω2 z(t)m 2L πx π2 dx = − 2 EAz(t) sin 4L 2L

(h)

The forces acting on a section of the bar (Fig. E2.7c) are seen to be in equilibrium. The result, which appears to be a coincidence, has been obtained because the shape function ψ(x) = sin(πx/2L) is one of the exact displacement mode shapes in which the bar can vibrate. The satisfaction of equilibrium does not, however, guarantee that the solution obtained is accurate. In fact, as we shall see later, the true vibration shape depends on the displacements and velocities imparted to the bar at time t = 0.

The procedure used to represent a distributed parameter system by a single-degreeof-freedom model illustrated by reference to the axial vibrations of a bar is equally applicable to a beam, a two-dimensional structure such as a flat plate, or to a threedimensional solid. In each case the displaced shape of the structure is expressed as the product of an appropriately selected shape function of the spatial coordinate(s)

Formulation of the equations of motion: Single-degree-of-freedom systems 47

multiplied by a generalized coordinate, which is a function of time and is the only unknown in the system. A virtual work equation is then written covering the work done by the inertia forces, the elastic forces, the forces of damping, and the applied forces when moving through a small admissible virtual displacement. An admissible displacement is one that is compatible with the constraints of the system. When the selected shape function of spatial coordinates satisfies these constraints, the virtual displacement is best chosen to be equal to the shape function multiplied by a variation δz of the generalized coordinate z. As a further example of the procedure, consider the beam shown in Figure 2.16a through d.The beam is supported on an elastic foundation with a spring constant of ¯ k(x) per unit length as well as by several local springs. It is subjected to a distributed ¯ force p(x) and one or more concentrated loads. The mass of the beam is m(x) ¯ per unit length. In addition, the beam supports several concentrated masses as shown. The damping forces are represented by viscous dampers distributed along the length and having damping constant c¯ (x) per unit length. Concentrated dampers are also specified. Let the displaced shape of the beam be represented by z(t)ψ(x), where ψ(x) is an appropriately chosen shape function and z(t) is the unknown generalized coordinate. The forces acting on a small section of the beam are indicated in Figure 2.16e. They include the reaction provided by the foundation, the distributed damping force, the applied load, the force of inertia, and the elastic moments acting on the ends of the section. The beam is now subjected to a virtual displacement given by δu = δz ψ(x). The relative rotation between the two faces of the section associated with the virtual displacement is δz ψ (x) dx. The virtual work done by the elastic moments is given by d(δWei ) = Mδz ψ (x) dx

(2.28)

Now, in accordance with the elementary beam theory, moment M is given by M = EI

∂2 u = EIz(t)ψ (x) ∂x2

(2.29)

Substituting Equation 2.29 in Equation 2.28, the virtual work done by the elastic forces becomes d (δWei ) = δz z(t)EI{ψ (x)} dx 2

(2.30)

As explained in Section 2.6, the internal virtual work d(δWi ) is the work done by the restoring elastic moments in the element and is negative of d(δWei ) given by Equation 2.30. It should also be noted that in calculating the virtual work of elastic forces, we have ignored the work done by the shear forces on true shear deformations. For normal proportions of the beam section, when the cross-sectional dimensions are small in comparison to the length of the beam, the work done by the shear forces is negligible in comparison to that done by the flexural forces. On adding the work done by the other forces indicated in Figure 2.16 to the internal virtual work done by the elastic moments and integrating over the length, we

48

Dynamics of structures

Figure 2.16 Vibrations of a beam.

obtain the following expression for the virtual work done by the distributed forces 

L

δW = −δz  + 0



0 L

L

2 m{ψ(x)} ¯ dx z¨ (t) +

2 ¯ k{ψ(x)} dx z(t) +

c¯ {ψ(x)}2 dx z˙ (t)

0



L 0

  2 EI{ψ (x)} dx z(t) + δz 0

L

¯ pψ(x) dx

(2.31)

Formulation of the equations of motion: Single-degree-of-freedom systems 49

The virtual work done by the concentrated masses is given by



δW = − δz

mi {ψ(xi )}2 z¨ (t) + δz



i

 I0i {ψ (xi )} z¨ (t) 2

(2.32)

i

where I0i is the mass moment of inertia of the ith mass, ψ(xi ) is the value of the shape function at the location of the ith mass, and ψ (xi ) is the value of the corresponding first derivative. If the concentrated masses are assumed to be point masses, the mass moment of inertia term drops out. In a similar manner, the virtual work of the concentrated spring forces is given by δW = − δz



 2

ki {ψ(xi )} z(t)

(2.33)

i

and that done by the concentrated dampers by δW = − δz



 2

ci {ψ(xi )} z˙ (t)

(2.34)

i

Finally, the virtual work done by the concentrated external forces is δW = δz



pi ψ(xi )

(2.35)

i

The total virtual work done by all the forces is obtained by summing Equations 2.31 through 2.35. The equation of motion is obtained by equating the total virtual work to zero (Eq. 2.9) m∗ z¨ (t) + c∗ z˙ (t) + k∗ z(t) = p∗

(2.36)

where the generalized mass m∗ , the generalized damping c∗ , the generalized stiffness k∗ , and the generalized force p∗ are given by

∗



L

m =

2 m{ψ(x)} ¯ dx +



0

c∗ =



i L

c¯ {ψ(x)}2 dx +

k∗ =

i L



2

EI ψ (x) dx +

0

p∗ =

 0

¯ pψ(x)dx +

i



2 I0i ψ (xi )

ci {ψ(xi )}2



L

2 ¯ k{ψ(x)} dx +

0 L



(2.37a)

i



0



mi {ψ(xi )}2 +

pi ψ(xi )

(2.37b)

ki {ψ(xi )}2

(2.37c)

i

(2.37d)

50

Dynamics of structures

Example 2.8 A uniform simply supported beam of mass m supports a point mass M at its center as shown in Figure E2.8. By selecting the deflected shape assumed by the beam under a central concentrated load as the shape function ψ(x), obtain the equation of motion.

Figure E2.8 Simply supported beam.

Solution The elastic curve is symmetrical about the center line of the beam. For x less than L/2 it is given by   x 2  x ¯ ψ(x) =C 3−4 x ≤ L/2 (a) L L where C is a constant. The constant C can be included in the generalized coordinate z(t), so that the shape function ψ(x) can be taken as (x/L) 3 − 4(x/L)2 . Also, ψ(L/2) = 1. The generalized mass is now given by m∗ = 2



L/2

m ¯

0


2 17 x2 17 x2 mL ¯ +M = m+M dx + M = 3 − 4 L2 L2 35 35

(b)

The generalized stiffness is given by k∗ = 2






L/2

EI 0

24x L3

2 dx =

48EI L3

The equation of motion is
 17 48EI m + M z¨ (t) + z(t) = 0 35 L3

(c)

(d)

It is interesting to note that with the foregoing idealization the uniform bar of mass m supporting a central mass M is entirely equivalent  to a massless bar having the same elastic properties but supporting a central mass of M + 17 m . 35

It is apparent from the foregoing discussion that the success of the procedure described for representing a distributed parameter system by a single-degree-offreedom model depends on an appropriate selection of the shape function ψ(x). This is often a difficult task because the vibration shape depends not only on the physical characteristics of the system but also on the type of loading. Ideally the selected shape

Formulation of the equations of motion: Single-degree-of-freedom systems 51

function should satisfy all boundary conditions of the system. These boundary conditions fall into two categories essential or geometric boundary conditions, and natural boundary conditions. As a minimum the shape function should satisfy the essential boundary conditions. As an example, consider a simply supported beam. The two essential boundary conditions specify that the deflection of the beam should be zero at each end. The natural boundary conditions prescribe zero moments at the two ends. The shape function selected in Example 2.8 satisfied all four conditions. Another shape function that would satisfy all four boundary conditions is ψ(x) = sin(πx/L). For the bar built-in at one end and free at the other and vibrating in the axial direction, the essential condition requires zero axial displacement at the built-in end, while the natural boundary condition calls for zero normal stress at the free end. The shape function ψ(x) = x/L satisfies the essential boundary condition but not the natural one. In addition to satisfying the essential boundary conditions, the shape function and its derivatives should satisfy certain conditions of continuity. From the examples presented here it is apparent that virtual work expressions for distributed parameter systems involve integration of the shape function and its derivatives. In the case of axial vibrations of a bar, for example, the highest-order derivative of the shape function appears in the expression for virtual work done by the elastic forces. The order of this derivative is 1. Apparently, the shape function should be once differentiable for the virtual work expression to be evaluated. This condition is satisfied if the function itself is continuous over the length of the bar. For flexural vibrations of a beam, the highest derivative appearing in the virtual work expression is of order 2. In this case, therefore, the shape function should be twice differentiable, implying that the first derivative of the shape function should be continuous over the length of the beam. In general, if the highest-order derivative in the virtual work expression is of order m, the (m − 1)th derivative of the shape function should be continuous over the domain. An alternative expression used to specify this condition is to state that the shape function should satisfy C m−1 continuity. The shape function method of modeling a distributed parameter system is useful only when the general nature of the vibration shape is known and a function that fits the shape can be selected. The difficulty in selecting a function increases with the dimensionality of the problem, and it may become impracticable to select a suitable function for a two-dimensional or a three-dimensional problem. In practical terms, therefore, the method has limited usefulness.

2.8

MODELING OF MULTI-DEGREE-OF-FREEDOM DISCRETE PARAMETER SYSTEM

In the preceding section we used an appropriately selected shape function to convert a distributed parameter system which has an infinite number of degrees of freedom to an equivalent single-degree-of-freedom system. A similar procedure can be used to represent a discrete parameter system having more than one degree of freedom by an equivalent single-degree-of-freedom system. As in the case of a distributed parameter system, the displaced shape of the discrete system is expressed as the product of

52

Dynamics of structures

Figure 2.17 (a) Three-story shear frame; (b) displaced shape; (c) forces acting along the degrees of freedom.

an appropriately selected shape function and a generalized coordinate. In this case, however, the shape function is a vector rather than a continuous function of the spatial coordinate(s). A virtual work equation is then written covering the work done by the inertia forces, the elastic forces, the forces of damping, and the applied forces in moving through a small virtual displacement. This virtual work equation directly gives the equation of motion. As an example of the procedure involved, consider the three-story shear frame shown in Figure 2.17a. The mass of the frame is assumed to be lumped at the floor levels. The floor beams are assumed to be rigid and the story stiffnesses are therefore provided by the flexure of the columns. The mass and stiffness properties are indicated in the figure. Damping resistance in the system is represented by interfloor dashpots. The shear frame shown in Figure 2.17a is constrained to vibrate in the plane of the paper and the columns are considered axially rigid. The frame has three degrees of freedom, as indicated. Each degree of freedom represents a possible lateral translation at a floor level. To model the frame by a single-degree-of-freedom system, we assume that the vibration shape is given by u = z(t)ψ, where ψ is a vector with three elements, representing the lateral translations of the floors, and z(t) is the unknown generalized coordinate. The inertia forces, the spring forces, and the damping forces acting at the floor levels are shown in Figure 2.17c. Also shown are the external forces applied at these levels. We now assume that the system is given a virtual displacement that is compatible with the constraints on the system. Such a virtual displacement can be represented by δz ψ. The virtual work equation becomes

Formulation of the equations of motion: Single-degree-of-freedom systems 53

(m3 z¨ ψ3 ) δz ψ3 + (m2 z¨ ψ2 ) δz ψ2 + (m1 z¨ ψ1 ) δz ψ1

+ k3 z(ψ3 − ψ2 ) δz ψ3 + k3 z(ψ2 − ψ3 ) + k2 z(ψ2 − ψ1 ) δz ψ2 + {k2 z(ψ1 − ψ2 ) + k1 zψ1 } δz ψ1 + {c3 z˙ (ψ3 − ψ2 )} δz ψ3 + {c3 z˙ (ψ2 − ψ3 ) + c2 z˙ (ψ2 − ψ1 )} δz ψ2 + {c2 z˙ (ψ1 − ψ2 ) + c1 z˙ ψ1 } δz ψ1 − p3 δz ψ3 − p2 δz ψ2 − p1 δz ψ1 = 0

(2.38)

Equation 2.38 can be written in a more compact form by using matrix notation: ψT Mψ¨z + ψT Cψ˙z + ψT Kψz = ψT p

(2.39)

where 

 m1 0 0 M = mass matrix =  0 m2 0  0 0 m3   c1 + c2 −c2 0 c2 + c3 −c3  C = damping matrix =  −c2 0 −c3 c3   k1 + k2 −k2 0 k2 + k3 −k3  K = stiffness matrix =  −k2 0 −k3 k3   p1 p = applied force vector =  p2  p3 Equation 2.39 can be expressed in the form m∗ z¨ + c∗ z˙ + k∗ z = p∗

(2.40)

where m∗ = ψT Mψ is the generalized mass, c∗ = ψT Cψ is the generalized damping, k∗ = ψT Kψ is the generalized stiffness, and p∗ = ψT p is the generalized force. As in the case of a distributed parameter system, the success of the method described here depends on how close the selected shape function is to the true vibration shape. Except in simple cases, a single shape function may not be sufficient to represent adequately the response of a multi-degree-of-freedom system.

2.9

EFFECT OF GRAV ITY LOAD

In formulating the equations of motion, we have in general ignored the presence of gravity loads. In this section we examine the effect gravity has on the equations of motion by reference to the simple system shown in Figure 2.18a. The system consists of a mass m suspended from the ceiling by a spring of stiffness k. The damping resistance is represented by a damper of stiffness c inserted between the ceiling and the

54

Dynamics of structures

Figure 2.18 Vibrating mass suspended from ceiling.

mass. The system is vibrating under the action of force p(t). When the system mass is at rest under its own load the spring will be stretched by an amount st given by mg = kst

(2.41)

Let the position of vibrating mass be specified by its distance ut from the end of the spring when the latter is unstretched. The forces acting on the mass are shown by the free-body diagram of Figure 2.18b. The equation of motion is readily obtained by applying the virtual work equation to the system shown in Figure 2.18b. mu¨ t + cu˙ t + kut = p(t) + mg

(2.42)

By substituting ut = u + st , where u is the displacement coordinate measured from the position the mass occupies when it is at rest under gravity load alone, Equation 2.42 becomes mu¨ + cu˙ + k(u + st ) = p(t) + mg

(2.43)

Substituting Equation 2.41 into Equation 2.43, we get mu¨ + cu˙ + ku = p(t)

(2.44)

The gravity load does not enter into Equation 2.44. This indicates that when the displacement coordinate is measured from the position of equilibrium under gravity, the forces arising due to gravity may be ignored in formulation of the equation of motion. Solution of such an equation will then give the relative displacement. If the total displacement is of interest, it can be obtained by adding the gravity

Formulation of the equations of motion: Single-degree-of-freedom systems 55

Figure 2.19 Simple pendulum.

load displacement to the relative displacement obtained by a solution of the dynamic problem. In the example just cited, the spring force consists of two components: the component kst which balances the gravity load mg, and the component ku, which opposes the vibrating motion. Further, the virtual work doneby the component kst exactly balances the virtual work done by the gravity force. The virtual work equation therefore does not contain any term involving the gravity force. This means that the gravity load can be omitted from the formulation provided that displacements are measured from the position of static equilibrium. The virtual work done by the gravity forces will not cancel out in every case, and when it does not, the gravity forces cannot be disregarded. In situations where the gravity forces must be considered, they act as either restoring forces or as destabilizing forces. Examples of each of the two cases are presented in the following paragraphs. A simple pendulum provides an example of the situation where the gravity load acts as a restoring force. As shown in Figure 2.19, the pendulum consists of a point mass m suspended by a light string of length l. When the point mass is displaced by an angle θ from its position of rest, the forces acting on it are as shown in the figure. To obtain the equation of motion, we give a small virtual displacement to the system. A displacement that is compatible with the constraints on the system should be along the tangent to the circular path described by the mass and can be represented by l δθ. The virtual work done by the forces of gravity is −mgl δθ sin θ. The virtual work performed by the inertia force is −ml 2 θ¨ δθ. The tension in the string that would balance the gravity force in the position of equilibrium does not perform any work. The equation of virtual work is (ml 2 θ¨ + mgl sin θ)δθ = 0

(2.45)

Canceling out ml δθ and noting that for vibrations of small amplitude sin θ = θ, we get the following equation of motion g θ¨ + θ = 0 l

(2.46)

56

Dynamics of structures

Figure 2.20 Inverted pendulum.

A situation where the gravity load acts as a destabilizing force exists in the case of an inverted pendulum, shown in Figure 2.20a. In the position of equilibrium, the gravity load mg is balanced by a thrust in the rod OA. The pendulum executes small vibrations about its position of equilibrium. The forces acting on the mass in its displaced position are as shown in Figure 2.20b. The equation of dynamic equilibrium can be obtained by taking moments about O. This gives ml 2 θ¨ + kl 2 θ cos θ − mgl sin θ = 0

(2.47)

For small θ, Equation 2.47 becomes  mg  mθ¨ + k − θ=0 l

(2.48)

The gravity load reduces the effective stiffness of the system. In fact, when m = kl/g, the system becomes unstable and buckles under its own weight. Example 2.9 The heavy flywheel shown in Figure E2.9 is rotating about a vertical axis at a constant angular speed of  rad/s. A small mass m is attached by a spring of stiffness k to the axle and is restrained by guides attached to the flywheel, so that the mass can move along the surface of the flywheel only in a radial direction. Obtain the equation of motion for radial vibrations of mass m.

Solution Let r be the unstretched length of spring. As the flywheel starts rotating, mass m moves away from the center of the wheel and eventually comes to rest at a distance x0 such that the spring force exactly balances the centrifugal force acting on the mass. Thus kx0 = m2 (r + x0 )

(a)

Formulation of the equations of motion: Single-degree-of-freedom systems 57

Figure E2.9 Mass moving on a rotating flywheel.

Suppose that the mass executes vibrations about the position of rest, its displacement from the position of rest being denoted by x. The forces acting on the mass at any instant of time are shown in Figure E2.9b and consist of the centrifugal force m2 (r + x0 + x), the inertia force mx, ¨ and the spring force k(x + x0 ). The equation of dynamic equilibrium becomes mx¨ − m2 (r + x0 + x) + k(x + x0 ) = 0

(b)

On substituting Equation a in Equation b, we get mx¨ + (k − m2 )x = 0

(c)

In this example the centrifugal force acts as a destabilizing force. When the speed of rotation is such that 2 = k/m, the system becomes unstable, since the spring is not stiff enough to restrain the mass from flying away off the wheel.

2.10 AXIAL FORCE EFFECT In Section 2.9 we presented an example where the gravity load acted as a destabilizing force, which effectively reduced the stiffness of the system. In Example 2.9, the destabilizing force was seen to be the centrifugal force. Axial forces acting in a system may also reduce its effective stiffness and may cause it to become unstable. Consider, for example, the simple rigid-body assemblage shown in Figure 2.21a. It consists of a rigid uniform bar of mass m and length L hinged at the left-hand end and supported at a distance b by a spring of stiffness k. The bar is subjected to a constant axial force S acting at its tip and is undergoing small vibrations in the plane of the paper. The forces acting on the bar when it is displaced by an angle θ about the hinge are indicated in Figure 2.21b. They include the inertia forces, the spring force, and the axial force. The reaction at the hinge is not shown. Also, thegravity forces have

58

Dynamics of structures

Figure 2.21 Axial force effect on a vibrating bar: (a) rigid bar; (b) free-body diagram; (c) virtual displacement.

been ignored; they do not affect the motion in this case. The equation of dynamic equilibrium is obtained quite simply by taking moments about the hinge. I0 θ¨ +

mL2 θ¨ + kb2 θ − SLθ = 0 4

(2.49)

Substituting I0 = mL2 /12 and setting u  Lθ, we get m u¨ + 3




 S kb2 − u=0 L2 L

(2.50)

or m∗ u¨ + (k∗ − kG )u = 0

(2.51)

where, m∗ = m/3, k∗ = kb2 /L2 , and kG = S/L. The axial load S has the effect of reducing the stiffness of the system by kG . The stiffness term kG is normally referred

Formulation of the equations of motion: Single-degree-of-freedom systems 59

Figure 2.22 Axial load on a section of a beam.

to as the geometric stiffness. When kG = k∗ , the system becomes unstable. The load at which instability will take place, called the buckling load, is given by S=

kb2 L

(2.52)

The equation of motion could also be obtained by writing the virtual work equation. The distance by which the axial load moves toward the hinge is given by y = L(1 − cos θ)

(2.53)

When the system is subjected to a virtual displacement δθ, the axial force undergoes a virtual displacement δy, where δy = L sin θ δθ

(2.54a)

which for small θ becomes δy = Lθ δθ

(2.54b)

The other virtual displacements are easily obtained and are shown in Figure 2.21c. The virtual work equation is given by −I0 θ¨ δθ −

mL2 θ¨ δθ − kb2 θ δθ + SLθ δθ = 0 4

(2.55)

On canceling out δθ, substituting I◦ = mL2 /12, and setting u ≡ Lθ, we get the equation of motion, which is identical to Equation 2.50. The effect of axial force on the motion of distributed parameter systems idealized as a single-degree-of-freedom system can be accounted for in a manner very similar to that described for the rigid-body assemblage described in the foregoing paragraphs. Consider, for example, the vibrations of the beam shown in Figure 2.16. The infinitesimal section of Figure 2.16e is redrawn in Figure 2.22 but with only the axial forces shown on it. The virtual displacement imposed on the beam will cause the section to rotate so that the forces S(x) shown in Figure 2.22 will move closer together by a

60

Dynamics of structures

distance δy which is obtained from d (1 − cos θ)δθ dx dθ = sin θ δθ dx

δy =

(2.56)

≈ θ δθ dx Since the displacement of the beam is u = zψ(x), the slope of the beam θ is given by θ = zψ (x)

(2.57)

The infinitesimal virtual work done by the axial force S(x) is now obtained from   (2.58) d(δWS ) = δz S(x){ψ (x)}2 dx z On integrating Equation 2.56 over the length of the beam, we get the following expression for the virtual work done by the axial force  δWS = δz

L

 S(x){ψ (x)}2 dx z

(2.59)

o

The geometric stiffness is now given by 

L

kG =

S(x){ψ (x)}2 dx

(2.60)

0

and the equation of motion is modified to m∗ z¨ (t) + c∗ z˙ (t) + (k∗ − kG )z(t) = p∗

(2.61)

For a static case, both z and p∗ are independent of time; hence z˙ (t) and z¨ (t) vanish. The resulting equation leads to an expression for deflection under static load and the condition k∗ = kG provides the value of axial load S(x) at which the system will buckle. In the special case of constant axial force, S can be taken out of the integral sign in Equation 2.58; the geometric stiffness is then given by  kG = S

L

{ψ (x)}2 dx

(2.62)

0

Example 2.10 The roof structure shown in Figure E2.10 consists of a uniform light column of flexural rigidity EI supporting a uniform reinforced concrete circular slab of radius R and mass m. Write the equation of motion for small vibrations of the structure in the plane of the paper (a) ignoring the axial force effect, and (b) including the axial force effect.

Solution If the column is considered axially rigid, the system has two degrees of freedom: a tip deflection, and a rotation, as shown in the figure. To represent the system by a

Formulation of the equations of motion: Single-degree-of-freedom systems 61

Figure E2.10 Vibrations of an umbrella roof.

single-degree-of-freedom system, a suitable displacement shape function must be selected. Let us assume that the displaced shape is given by  πx  u = z 1 − cos 2L where z is the generalized coordinate and ψ(x) = 1 − cos(πx/2L) is the displacement shape function. (a)

The generalized mass m∗ is obtained from Equation 2.37a: m∗ = m{ψ(L)}2 + I0 {ψ (L)}2 where I0 = mR2 /4 is the mass moment of inertia of the roof slab about its diameter. Noting that ψ(L) = 1 and ψ (L) = π/2L, we get mR2  π 2 4 2L
2  2 R π =m 1+ 16 L

m∗ = m + 

(a)

The generalized stiffness is obtained from Equation 2.37c: k∗ =



L



2 EI ψ (x) dx

0

 =

L

EI 0

=

π4 EI 32L3

π4  πx 2 dx cos 16L4 2L

(b)

62

Dynamics of structures The equation of motion becomes 

π2 m 1+ 16 (b)


2  R π4 EI z=0 z¨ + L 32L3

(c)

The weight of the roof slab imposes an axial force mg on the column. The geometric stiffness is obtained from Equation 2.62  kG = mg 0

L

π2 πx dx sin2 4L2 2L

mgπ2 = 8L

(d)

The modified equation of motion becomes 

π2 m 1+ 16


2  
4 R mgπ2 π EI z=0 − z¨ + L 32L3 8L

(e)

The column supporting the roof will buckle under the weight of the roof when k∗ = kG , that is, when mg =

π2 EI 4L2

(f)

The buckling load given by Equation f is the true buckling load of a uniform vertical cantilever loaded at its end. This happens to be so because the assumed displacement shape is, in fact, the true displaced shape at buckling.

2.11

EFFECT OF SUPPORT MOTION

Dynamical systems are often excited by the motion of their supports. Earthquakeinduced motion provides one example of support excitation. For a single-degree-offreedom system the equation governing a support excited motion is easily obtained. A system of this type can be modeled as shown in Figure 2.23a. Let ut denote the displacement of the mass with reference to a fixed frame in space and ug denote the displacement of the ground with reference to such a frame. The displacement of the mass relative to the ground is represented by u so that ut = ug + u

(2.63)

The forces acting on the moving mass are shown in the free-body diagram of Figure 2.23b. It should be noted that the inertia force is given by the product of the mass and the absolute acceleration relative to the fixed frame. The spring and damping forces, on the other hand, depend on the displacement and velocity, respectively, relative to the ground. The equation of motion for the system is given by mu¨ t + cu˙ + ku = 0

(2.64)

Formulation of the equations of motion: Single-degree-of-freedom systems 63

Figure 2.23 Effect of support motion.

On substituting for ut from Equation 2.63, Equation 2.64 becomes mu¨ + cu˙ + ku = −mu¨ g

(2.65)

Equation 2.63 expresses the equation of motion in terms of the relative displacement coordinate and its derivatives, with the term −mu¨ g , in effect, acting as an exciting force.

SELECTED READINGS Best, C.W. “Material Damping: An Introductory Review of Mathematical Models, Measures and Experimental Techniques’’, Journal of Sound and Vibration, Vol. 29, 1973, pp. 129–153. Clough, R.W., and Penzien, J., Dynamics of Structures, 2nd Edition, McGraw-Hill, New York, 1993. Crandall, S.H., “The Role of Damping in Vibration Theory’’, Journal of Sound and Vibration, Vol. 11, 1970, pp. 3–18. Fowles, G.R., Analytical Mechanics, 4th Edition, Holt, Rinehart and Winston, New York, 1986. Lazan, B.J., Damping of Materials and Members in Structural Mechanics, Pergamon Press, Oxford U.K., 1968. Nashif, A.D., Jones, D.I.G. and Henderson, J.P., Vibration Damping, John Wiley, New York, 1985. Pinsker, W., “Structural Damping’’, Journal of the Aeronautical Sciences, Vol. 16, 1949, pp. 619. Scanlan, R.H. and Mendelson, A., “Structural Damping’’, AIAA Journal, Vol. 1, 1963, pp. 938–939.

PROBLEMS 2.1

Design an arrangement of three springs with stiffness 40, 60, and 90 N/mm, respectively, to have an effective stiffness of 76 N/mm.

64

Dynamics of structures

2.2

Derive expressions for the resultant inertia forces for translation along the x, and y axes, and rotation about the mass center for the rigid bodies shown in Figures 2.6c and d. In each case assume that the body has a uniform thickness and mass density.

2.3

A rigid block of mass M is suspended from a uniform pulley of mass m and radius R as shown in Figure P2.3. If the cord is inextensible, of negligible weight, and does not slip on the pulley, and the spring stiffness is k, determine the equation of motion for vertical vibrations of the system.

Figure P2.3

2.4

A wheel-axle assembly weighing 10 lb is mounted at the end of a cantilever strip of length 1 ft, width 1 in., and thickness 1/4 in. (Fig. P2.4). The strip is rigidly attached to a solid circular shaft of length 3 ft and diameter 1 in. The other end of the shaft is rigidly fixed to a heavy vehicular body. The wheel-axle assembly has a radius of gyration of 3.5 in. E = 30 × 106 psi, and G = 12 × 106 psi.

Figure P2.4

Obtain the equation of motion for vertical vibrations of the wheel-axle assembly assuming that the wheel is free to rotate about the axle and the friction in the bearings is negligible.

Formulation of the equations of motion: Single-degree-of-freedom systems 65 2.5

Obtain the equation of motion for small vibrations of the system shown in Figure P2.5.

Figure P2.5

2.6

A uniform flywheel of mass M and radius R rotates freely about an axle carried by two identical rigid bars of total mass m as shown in Figure P2.6. The rigid bars are pinned at their left-hand end and suspended by a vertical spring at the right-hand end. A dashpot with damping coefficient c is attached to the bars at a distance a from the pinned end. Assuming that there is no friction in the bearings of the wheel, obtain the equation of motion for small vertical vibrations of the system. What will be the equation of motion, if the flywheel was braked to its axle.

Figure P2.6 2.7

Shrouds are often used at the tips of turbine blades to improve performance and to control vibration of the blades (Fig. P2.7). For purposes of vibration analysis, the shrouded blade can be considered to behave as a cantilevered beam with the tip pinned.

66

Dynamics of structures

Figure P2.7

In order to model the blade as a single-degree-of-freedom system, the following function is used to represent its vibration shape: ψ(ξ) = 2ξ 4 − 5ξ 3 + 3ξ 2 , (a) (b)

2.8

ξ=

x L

To what extent does ψ(ξ) meet the boundary conditions? Determine the generalized mass m∗ and the generalized stiffness k∗ and hence obtain the equation of free vibration of the blade. Assume that the blade is of uniform section, the mass per unit length is m, ¯ and the flexural rigidity is EI.

Ultrasonic transducers often contain a “transformer’’ which has the purpose of magnifying the input vibration amplitude as shown in Figure P2.8. Assuming the shape function for axial vibration as ψ(x) = 2ξ − ξ 2 , where ξ = Lx and that the cross-sectional area varies linearly so that A(ξ) = A1 + (A2 − A1 )ξ, obtain expressions for m∗ and k∗ . If the input is a base motion given by ug (t), formulate the equation of motion.

Figure P2.8 2.9

A vertical chimney of length L (Fig. P2.9) has a uniform cross-section, moment of inertia I, and mass per unit length m. ¯ The modulus of elasticity of the material is E. The chimney is subjected to lateral ground motion having acceleration u¨ g (t). Find the equation of motion: (a) neglecting the gravity load effect produced by the self weight of the chimney; (b) taking gravity load effect into account. Ignore damping and assume that the vibration shape is given by: ψ(x) =

3x2 x3 − 2 2L 2L3

Formulation of the equations of motion: Single-degree-of-freedom systems 67

Figure P2.9

2.10

For the simplified analysis of the response of a bridge to moving loads, the bridge deck is idealized as a simply supported beam of span L, mass per unit length m, ¯ and flexural rigidity EI. A single wheel load of magnitude F traverses the bridge at a speed v (Fig. P2.10). Assuming a displacement shape function given by ψ = sin(πx/L), obtain the equation of motion for flexural vibrations of the bridge deck. Ignore the mass of the wheel.

Figure P2.10

2.11

The following properties are given for the frame shown in Figure 2.17: m1 = m2 = 2 2 kip s /in., m3 = 1 kip s2 /in., k1 = k2 = 500 kips/in., and k3 = 250 kips/in. The frame is subjected to lateral ground motion whose acceleration varies as u¨ g = 130 sin(6πt) in./s2 . Obtain the equations governing the motion of the frame. Ignore damping and assume that the vibration shape is given by:  ψT = 1

2

3



68 2.12

Dynamics of structures A mass m is attached to a massless rigid bar of length l. The bar is suspended from a ceiling and is restrained by two springs each of stiffness k, as shown in Figure P2.12. Find the equation of motion for small vibrations of the bar in a vertical plane.

Figure P2.12 If the whole assembly were in a horizontal plane, what would be the equation of motion.

Chapter 3

Formulation of the equations of motion: Multi-degree-of-freedom systems

3.1

INTRODUCTION

For an accurate description of its displaced configuration, a structural or mechanical system subjected to dynamic disturbances may require the specification of displacements along more than one coordinate direction. Such a system is known as a multi-degree-of-freedom system. In this chapter we describe procedures for formulation of the equations of motion of multi-degree-of-freedom systems. These procedures are similar in principle to those used for the single-degree-of-freedom systems, even though, in detail, they are quite a bit more involved. In Chapter 2 we discussed a method by which the vibrations of a multi-degreeof-freedom system could be adequately represented by a single shape function. Such a representation is, in fact, an idealization that is only occasionally valid. In a general case, accurate description of the displaced configuration of a vibrating system is possible only through the superposition of a number of different shapes. Even when the use of a single shape function is adequate, the selection of the shape function is not, in general, easy, and if an inappropriate choice is made, the results obtained may be completely unreliable. The difficulty is compounded by the fact that in procedures that use shape function idealization, there is no simple way to verify the reliability of the results obtained. As a simple example of the difficulties involved in using a single-degree-of-freedom idealization of what is truly a multi-degree-of-freedom system, consider the two-mass system shown in Figure 3.1. To describe the displaced configuration of the system completely, we need to specify the displacements along two coordinate directions: u1 and u2 . We may attempt to idealize the system as a single-degree-of-freedom system by choosing a shape function ψ such that ψT = [1

1]

(3.1)

With this idealization, the response of the system is given by u = z(t)ψ

(3.2)

The only unknown in Equation 3.2 is z(t) and the shape function assumption has allowed us to represent a two-degree-of-freedom system by a single-degree-of-freedom model.

70

Dynamics of structures

Figure 3.1 Vibrations of a two-degree-of-freedom system.

Now let the system be excited from rest by the application of two equal forces, each equal to p(t), and acting in the same direction, as indicated in Figure 3.1a. By using the method described in Section 2.8, we obtain the following equation for the single-degree-of-freedom idealization of the system: m∗ z¨ + k∗ z = p∗

(3.3)

where m∗ = 2m, k∗ = 2k, and p∗ = 2p. In the present case, the solution of Equation 3.3 gives the exact response. However, let us now assume that instead of acting in the same direction, the two equal forces act in opposite directions, as in Figure 3.1b. For such forces, p∗ in Equation 3.3 will be equal to zero and that equation will entirely fail to predict the response of the system. The problem is that the shape function ψT = [1 1] is no longer appropriate; instead, we must use ψT = [1 −1]. In fact, for the general loading shown in Figure 3.1c, a single shape function will not be adequate to provide the response of the system; a superposition of two shape functions must be used. Any number of examples can be cited of a multi-degree-of-freedom system, but consider the simple flexible cantilever beam of Figure 3.2 supporting three lumped masses. First, let the masses be point masses and let the beam be constrained to vibrate in the plane of the paper. If the cantilever is axially rigid, the masses can undergo vibrations only in the vertical direction and the system has three degrees of freedom. If the beam is axially flexible, horizontal motion of the masses is also possible, making the system a six-degree-of-freedom system. Next, if the masses are of finite size instead

Formulation of the equations of motion: Multi-degree-of-freedom systems 71

Figure 3.2 Cantilever beam with lumped masses.

of being concentrated at a point, they have rotational inertia, too, and three more rotational degrees of freedom must be specified, one for each mass. In the most general case, when the cantilever can undergo motion in a three-dimensional space, each mass has six degrees of freedom, three translational, and three rotational, and the cantilever has a total of 18 degrees of freedom. In conclusion, we may state that it is essential in many cases for the system being analyzed to be represented by a multi-degree-of-freedom system. In this chapter we present techniques for the formulation of the equations of motion for such systems.

3.2

PRINCIPAL FORCES IN MULTI-DEGREE-OF-FREEDOM DYNAMIC SYSTEM

As in the case of a single-degree-of-freedom system, the principal forces acting on a multi-degree-of-freedom system are (1) the forces of inertia, (2) the elastic forces, and (3) the forces of resistance or damping. Once these forces have been identified and calculated, formulation of the equations of motion reduces to a problem of writing the equations of equilibrium. In this section we describe methods of obtaining these forces.

3.2.1 Inertia forces Consider the lumped mass idealization of a simply supported beam shown in Figure 3.3. The model consists of N point masses which are free to vibrate in the vertical direction in the plane of the paper. To define the displaced configuration of this system, we need to specify the displacements along N coordinates. For example, if we know the vertical displacement at each point mass, the displaced shape of the beam is uniquely determined. The system is therefore an N-degree-of-freedom system and we may choose the vertical displacements at the point masses as the N coordinates. Next, let us imagine that the system has a unit acceleration along one of the coordinate directions, say j, while the accelerations at all other coordinates are zero. Since the coordinates are independent of each other, such a state is possible. For our choice of coordinates, this state implies that only the mass Mj is undergoing acceleration, and therefore the force of magnitude −Mj · 1 shown is the only inertia force in the system. The negative sign implies that the inertia force is directed away from the direction

72

Dynamics of structures

Figure 3.3 Inertia forces in a lumped mass model of a simply supported beam.

of the positive unit acceleration. We now determine the external forces that must be applied along the N coordinate directions to equilibrate the inertia force. Denoting by mij the force that must be applied along coordinate i to equilibrate the inertia forces produced by a unit acceleration at coordinate j, we note that mij = 0

i = j

(3.4)

mjj = Mj

By superposition of the forces it can easily be shown that the external force vector required to equilibrate the inertia forces arising from simultaneous acceleration of all masses, the mass accelerations being u¨ j , j = 1 to N, is given by   m11 fI1  fI2   m21     ·   ·   fI =   · = ·     ·   · fIN mN1 

fI = Mu¨

m12 m22 · · · mN2

  · · · m1N u¨ 1  u¨ 2  · · · m2N      · · · ·   ·    · · · ·   ·    · · · · ·  u¨ N · · · mNN

(3.5a)

(3.5b)

The matrix on the right-hand side of Equation 3.5a is called the mass matrix and is denoted by M. Individual elements mij of the mass matrix are referred to as the mass influence coefficients. In our particular case, because of relationship in Equation 3.4, the mass matrix is diagonal, the terms on the diagonal being Mj , j = 1 to N. It will readily be appreciated that the mass matrix need not always be diagonal. As an example, consider the rigid bar shown in Figure 3.4a. The bar, which is nonuniform, has a total mass m and is supported at its two ends by massless springs of stiffness k1 and k2 . The mass center of the bar is at a distance a from the left-hand end, and the mass moment of inertia about this center is I0 . The bar is constrained so that it can

Formulation of the equations of motion: Multi-degree-of-freedom systems 73

Figure 3.4 Vibrations of a rigid bar, inertia forces.

move only in a vertical direction in the plane of the paper. In order that the position of the bar be completely determined, we need to specify the displacements along two coordinates. There are a number of possible choices for the coordinates; we choose the vertical displacements at the bar ends as our two coordinates. To obtain the inertia forces acting on the bar, we first apply a unit acceleration along coordinate 1, the acceleration at the other coordinate being zero. This will cause a translational acceleration of b/L at the center of mass along with a clockwise rotational acceleration of 1/L about the same center. The resulting inertia forces acting at the mass center are indicated in Figure 3.4b. They consist of a vertical force mb/L directed downward and an anticlockwise moment I0 /L. If the bar is to be kept in dynamic equilibrium, the inertia forces must be balanced by appropriate forces applied at each of the two coordinate directions. Using the notation already introduced, the force at coordinate 1 is denoted by m11 , being the external force required at coordinate 1 to equilibrate the inertia forces arising from the application of a unit acceleration also at coordinate 1. In a similar manner, the force at coordinate 2 is denoted by m21 , being the external force required at coordinate 2 to equilibrate the inertia forces arising from the application of a unit acceleration at coordinate 1. The magnitudes of m11 and m21 are easily obtained by considering the equilibrium of the bar. Thus taking

74

Dynamics of structures

moments about the right-hand end yields 1 b m11 L = m b + I0 L L I0 mb2 m11 = 2 + 2 L L

(3.6)

In a similar manner, m21 =

mab I0 − 2 L2 L

(3.7)

Next, we apply a unit acceleration at coordinate 2 with the acceleration at coordinate 1 being zero. The resulting inertia forces at the center of mass are indicated in Figure 3.4c. The external forces required to equilibrate the inertia forces are denoted by m12 and m22 , respectively, and are given by mab I0 − 2 2 L L 2 ma I0 m22 = 2 + 2 L L

m12 =

(3.8) (3.9)

If the actual accelerations in the system are u¨ 1 and u¨ 2 , the external forces required to balance the resulting inertia forces are easily seen to be      m11 m12 u¨ 1 fI1 = fI2 m21 m22 u¨ 2   2 mb + LI02 mab − LI02 u¨ 1 L2 L2 (3.10) = I0 I0 mab ma2 u¨ 2 − + 2 2 2 2 L L L L or fI = Mu¨

(3.11)

Several interesting facts can be noted from Equation 3.10. First, the mass matrix is no longer diagonal; that is, the-off diagonal terms have nonzero values. Second, the matrix is symmetric, that is, m12 = m21 . The latter is a direct consequence of Maxwell’s reciprocal theorem and, in general, mij = mji .

3.2.2 Forces arising due to elasticity As in the case of a single-degree-of-freedom system, whenever an elastic body is subjected to deformations, internal forces are set up that counter such deformations. Therefore, if the deformations are to be maintained, external forces must be applied along the coordinate directions of the system. We will refer to these external forces as the spring forces or the forces arising from elasticity; in fact, they are forces that must be applied to counter the internal elastic forces.

Formulation of the equations of motion: Multi-degree-of-freedom systems 75

Figure 3.5 Lumped mass model of a beam: (a) imposed displacement pattern; (b) elastic forces required to maintain displacement in (a).

When the system is linear, that is, when the stress–strain relationship of the material used in the structure is linear and the displacements are small, the elastic forces can be obtained by the method of superposition. To illustrate the procedure, we consider again the example of lumped mass beam model shown in Figure 3.3. We apply a unit displacement along coordinate direction j, holding all other displacements to zero as shown in Figure 3.5a. Internal elastic forces will oppose these displacements, and to maintain the displacements we must apply external forces along all coordinate directions as indicated in Figure 3.5b. We denote the external force required at coordinate i as kij . If the actual displacements in the system are uj , j = 1 to N, it is easily shown using the principle of superposition that the external forces along the coordinate directions required to balance the internal elastic forces are given by      k11 k12 · · · k1N u1 fS1  fS2   k21 k22 · · · k2N   u2          ·   · · · · · ·  =  ·  fS =  (3.12)    ·   · · · · · ·     ·    ·   · · · · · ·  ·  fSN kN1 kN2 · · · kNN uN or fS = Ku

(3.13)

where the matrix on the right-hand side of Equation 3.12 is referred to as the stiffness matrix and is denoted by K. Individual elements kij of the stiffness matrix are referred to as the stiffness influence coefficients. As in the case of mass matrix, the stiffness matrix is also symmetrical. As another example, we derive the spring forces for the rigid bar shown in Figure 3.4. For the coordinate directions selected, the spring forces are shown in Figure 3.6, and are given by    fS1 k1 0 u1 = (3.14) 0 k2 fS2 u2 Thus for the coordinates selected, the stiffness matrix is diagonal.

76

Dynamics of structures

Figure 3.6 Vibrations of a rigid bar, elastic forces.

3.2.3 Damping forces As pointed out in Section 2.5, the motion of a body is opposed by several types of resisting forces. These forces may, for example, arise from air resistance or from internal and external frictions. As noted earlier, the characteristics of the resisting forces are difficult to define. From a mathematical point of view, viscous damping forces which are proportional to the velocities in the system but are opposed to the direction of motion are easiest to handle and give analytical results that for small amounts of damping conform reasonably well to experimental observations. In this section we deal only with viscous damping forces. Later, we will have occasion to discuss other types of damping forces. Consider again the lumped mass model of a simply supported beam. Let us impart a unit velocity at coordinate direction j, the velocities at other coordinates being zero (Fig. 3.7a). The specified motion will be opposed by damping forces in the system, and if the motion is to be maintained, external forces as shown in Figure 3.7b must be applied at the coordinate directions to balance the forces of resistance. As in the case of inertia and spring forces, we denote the external force at coordinate i by cij . If the real velocities along the coordinates are u˙ j , j = 1 to N, the damping forces are obtained by superposition. Thus      c11 c12 · · · c1N u˙ 1 fD1  fD2   c21 c22 · · · c2N   u˙ 2          ·   · · · · · ·     ·   fD =  (3.15) = ·  ·  · · · · · ·       ·   · · · · · ·  ·  fDN cN1 cN2 · · · cNN u˙ N

Formulation of the equations of motion: Multi-degree-of-freedom systems 77

Figure 3.7 Lumped mass model of a beam: (a) velocity pattern; (b) external forces required to balance resistance forces caused by damping; (c) damping model.

or fD = Cu˙

(3.16)

where the matrix on the right-hand side of Equation 3.15 is called the damping matrix and is denoted by C. Elements cij of the damping matrix are referred to as the damping influence coefficients. The damping influence coefficients cij are similar in concept to the mass influence coefficients mij and the stiffness influence coefficients kij . We could derive influence coefficient mij ’s from a knowledge of the internal mass distribution in the system. In a similar manner, kij ’s could be derived from the internal stiffness characteristics or the stress–strain relationship of the material. However, internal damping characteristics are difficult or impossible to define, and therefore the coefficients cij ’s can rarely be obtained from the consideration of internal damping characteristics. In later chapters we shall discuss alternative methods of constructing damping matrices or defining damping resistances. For the present we assume that cij ’s have been specified or can be computed. As a specific example, consider the beam model with the damping characteristics shown in Figure 3.7c. For this particular model cij = 0

i = j

cii = Ci and the damping matrix is diagonal.

(3.17)

78

Dynamics of structures

3.2.4 Axial force effects As in the case of a single-degree-of-freedom system, the presence of axial forces usually causes a reduction in the resistance that the system offers to elastic deformations. The axial force effect can be represented by a force vector fG that is opposite in direction to the spring force vector fS . Consider, for example, the lumped mass model of a beam. Suppose that we wish to impose a unit displacement along coordinate direction j while holding to zero the displacements along other coordinate directions. As stated earlier, the resulting deformations will be resisted by internal elastic forces. At the same time, the presence of axial forces will tend to amplify the deformations. The net external force required along coordinate direction i to maintain the displacements will be kij − kGij , where kij is the elastic stiffness influence coefficient, discussed earlier, and kGij is the geometric stiffness influence coefficient. If the displacements along the coordinates are uj , j = 1 to N, the forces of instability that tend to amplify the deformations are given by   kG11 fG1  fG2   k   G21  fG =  .  =  .  ..    .. fGN kGN1 

kG12

···

kG1N

kG22 .. .

··· .. .

kG2N .. .

kGN2

···

kGNN



 u1     u2   .   .   . uN

(3.18)

or fG = KG u

(3.19)

in which KG is referred to as the geometric stiffness matrix. As an example of the forces of instability, consider the rigid bar of Figure 3.6. Let the bar be subjected to an axial force S. Figure 3.8a shows the displaced shape of the bar when a unit displacement has been imposed along coordinate direction 1, the displacement along coordinate 2 being zero. The resulting internal spring forces and

Figure 3.8 Axial force effect on the vibration of a rigid bar.

Formulation of the equations of motion: Multi-degree-of-freedom systems 79

the external forces required to counter them were derived earlier and are not shown. However, the axial forces S give rise to a couple S × 1. To maintain equilibrium, this couple must be balanced by external forces kG 11 and kG 21 acting along coordinate directions 1 and 2 as shown. Using equations of static equilibrium, we get kG 11 =

S L

(3.20)

S kG 21 = − L

In a similar manner, by imposing a unit displacement along coordinate 2, as shown in Figure 3.8b, we get kG 12 = −

S L

(3.21)

S kG 22 = L

If the real displacements along the coordinate are u1 and u2 , the total forces of instability are given by fG =

fG1 fG2



=

S L − LS

− LS



S L

u1



u2

= KG u

(3.22)

It should be noted that the signs of the geometric stiffness influence coefficients have been determined with the understanding that fG will be deducted from fS .

3.3

FORMULATION OF THE EQUATIONS OF MOTION

Once the external forces required to equilibrate the inertia forces, damping forces, the elastic forces, and the forces of instability have been obtained by the procedures discussed in Section 3.2, the formulation of the equations of motion is straightforward. For equilibrium, the sum of the above-mentioned forces should be equal to the actual external forces acting along the coordinates of the system. If these external forces are denoted by p(t), the equations of motion become fI + fD + fS − fG = p(t)

(3.23)

Mu¨ + Cu˙ + Ku − KG u = p(t)

(3.24)

or

When the external forces in the system act at locations other than the coordinates defined for the system, the former must be replaced by equivalent forces acting at the

80

Dynamics of structures

coordinates. As discussed in succeeding paragraphs, this is usually best accomplished by using the principle of virtual displacement. In dealing with more specific examples of the formulation of the equations of motion for multi-degree-of-freedom systems, we will find it convenient to classify the latter into one of the following categories, just as we did in the case of single-degreeof-freedom systems. 1 2 3 4

Systems with localized mass and localized stiffness Systems with localized mass but distributed stiffness Systems with distributed mass but localized stiffness Systems with distributed mass and distributed stiffness

3.3.1 Systems with localized mass and localized stiffness Figure 3.9 shows a system consisting of two rigid blocks restrained by springs and vibrating in a horizontal direction in the plane of the paper. The forces of elastic constraints are localized in the massless springs, while the masses can be assumed to be concentrated at the mass centers of the rigid blocks. The system, which has two degrees of freedom, identified in Figure 3.9 by the two displacement coordinates, is vibrating under the action of externally applied forces p1 and p2 , also shown in Figure 3.9. Using the procedure outlined in Section 3.2, the mass, stiffness, and damping matrices are easily obtained and are given by  m1 0 M= (3.25a) 0 m2  c1 + c2 −c2 C= (3.25b) −c2 c2  k1 + k2 −k2 K= (3.25c) −k2 k2 The equations of motion, which are of the form of Equation 3.24, can now be written as            m1 0 u¨ 1 c1 + c2 −c2 u˙ 1 k1 + k2 −k2 u1 p1 + + = 0 m2 u¨ 2 −c2 c2 u˙ 2 −k2 k2 u2 p2 (3.26)

Figure 3.9 System with localized mass and localized stiffness.

Formulation of the equations of motion: Multi-degree-of-freedom systems 81

3.3.2 Systems with localized mass but distributed stiffness Figure 3.10a shows a simply supported massless beam, with point masses concentrated at one-third points along the span. The beam is axially rigid and is constrained to move in the plane of the paper. The system shown in Figure 3.10a can, in fact, be viewed as a lumped mass model of a simply supported uniform beam whose mass has been lumped at one-third points. The beam is vibrating under a central load p0 (t), and it is required to obtain the equations of motion. The given beam can be classified as a system having localized masses but distributed stiffness. It has two degrees of freedom corresponding to the vertical translation of each mass as shown in the figure. The inertia forces act along these two degrees of freedom. The external force does not act through the selected degrees of freedom, but can be replaced by equivalent forces acting through these degrees of freedom by using the principle of virtual displacement as discussed later in this section. The forces of elastic restraint are distributed throughout the length of the beam but can be replaced by resultant forces acting along the two degrees of freedom using standard methods of structural analysis. Once the response along the two selected degrees of freedom has been determined, the deformed shape of the beam can be determined by methods of elastic analysis. The mass matrix of the beam model is given by  M=



m m

(3.27)

Instead of calculating the stiffness matrix of the beam directly, it is more convenient first to obtain the flexibility matrix of the beam in the coordinates shown and then to take the inverse of this matrix.

Figure 3.10 (a) Lumped mass model of a simply supported beam; (b), (c) flexibility influence coefficients.

82

Dynamics of structures

The flexibility influence coefficient aij is defined as the displacement at coordinate i due to a unit load acting along coordinate j. Thus to obtain a11 and a21 , we apply a unit load at coordinate 1 and compute the deflections produced due to this load at coordinates 1 and 2, respectively (Figure 3.10b). These deflections are easily obtained by standard procedures of structural analysis and lead to the following values for the influence coefficients: a11 = a21

4L3 243EI

(3.28)

7L3 = 486EI

In a similar manner, to obtain a12 and a22 , we apply a unit load at coordinate 2 and compute the deflections produced at coordinates 1 and 2 respectively, (Fig. 3.10c). This gives a12 = a22

7L3 486EI

(3.29)

4L3 = 243EI

The off-diagonal elements a12 and a21 are equal, as would be expected on account of the Maxwell’s reciprocal theorem. Coefficients a11 and a22 are also equal in this special case because of symmetry. The flexibility matrix is now given by L3 A= 243EI



4

7 2

7 2

4

 (3.30)

The stiffness matrix is obtained by taking the inverse of the flexibility matrix: K = A−1

324 EI = 5 L3



4

− 72

− 72

4

 (3.31)

The principle of virtual displacement is used to obtain the equivalent applied force at the two coordinates. Let a virtual displacement δu1 be applied at coordinate 1 while the displacement at coordinate 2 is zero. Let the resulting displacement at the load point be 10 . The virtual work equation then gives p1 δu1 = p0 10

(3.32)

where p1 is the equivalent force at coordinate 1. To obtain 10 , we first recognize that the forces F1 and F2 that must act at coordinates 1 and 2 to produce the displacement

Formulation of the equations of motion: Multi-degree-of-freedom systems 83

pattern δu1 , 0 are given by      324EI 4 − 72 δu1 F1 = F2 0 4 5L3 − 72  1296 EI δu1 5 L3 = EI − 2268 δu1 10 L3

(3.33)

The deflection produced at midspan due to the set of forces F1 and F2 is obtained by standard methods of structural analysis and is 10 =

23 δu1 40

(3.34)

Substitution of Equation 3.34 in Equation 3.32 gives p1 =

23 p0 40

(3.35)

In a similar manner, it can be shown that p2 =

23 p0 40

(3.36)

The equivalent loads p1 and p2 are often referred to as consistent loads, and it is interesting to note that in this case, they are not statically equivalent to the applied load. Often, however, static equivalents of the applied load provide a reasonable approximation of the load effect and are used in place of consistent loads. In our example, the static equivalents will be p1 = p2 = p/2. Using the load vector and ignoring damping, the equations of motion can be written as      23 u1 m u¨ 1 4 − 72 324EI 40 + = p (3.37) 0 23 5L3 m u¨ 2 u2 4 − 72 40

3.3.3 Systems with distributed mass but localized stiffness The rigid bar of Figure 3.4 is an example of a system that can be categorized as having a distributed mass but localized stiffnesses. It is redrawn in Figure 3.11a, where the external forces consisting of a direct force and a moment are also shown. It is assumed that the damping is negligible. The spring forces on the bar are concentrated at the two locations where the springs are attached to the bar. However the inertia forces are distributed throughout the length. In spite of this the system has only two degrees of freedom, because we can completely define the displaced position of the bar in terms of the displacements along two selected degrees of freedom. An infinite number of choices exist when selecting the two degrees of freedom. In the present case we select vertical displacements at the

84

Dynamics of structures

Figure 3.11 Vibrations of a rigid bar: (a) applied loads; (b) equivalent loads.

two ends of the bar, u1 and u2 , as shown in Figure 3.11. The displaced shape of the bar is related to u1 and u2 as follows: u(x, t) = u1 (t)ψ1 (x) + u2 (t)ψ2 (x)

(3.38)

where u is the vertical displacement of the bar at a distance x from the left-hand end and the shape functions ψ1 (x) and ψ2 (x) are given by ψ1 (x) = 1 − ψ2 (x) =

x L

x L

(3.39a) (3.39b)

The two shape functions are shown by dashed lines in Figure 3.11b, and are exact because the bar is rigid. As in the case of a single-degree-of-freedom system, coordinates u1 and u2 are known as generalized coordinates. The mass matrix of the system corresponding to the two-degree-of-freedom system shown in Figure 3.11a was obtained in Section 3.2.1 (Eq. 3.10). The stiffness matrix for the same two-degree-of-freedom system is given by Equation 3.14. The externally applied forces do not act along the coordinate directions. Therefore, before formulating the equations of motion, we must convert the applied forces into equivalent forces along the coordinates. These equivalent forces are identified as P1 and P2 in Figure 3.11b.

Formulation of the equations of motion: Multi-degree-of-freedom systems 85

The principle of virtual work can be effectively utilized to calculate the magnitude of forces P1 and P2 . Thus if we apply a virtual displacement δu1 along coordinate direction 1, the work done by the applied forces is given by δu1 b W = F δu1 − M L L

(3.40)

On the other hand, the virtual work done by the equivalent forces is given by WE = P1 δu1 + P2 × 0

(3.41)

Because the work obtained by the two alternative methods should be the same, we equate Equations 3.40 and 3.41 and get P1 = F

b M − L L

(3.42)

In a similar manner, by applying a virtual displacement δu2 , we obtain P2 = F

M a + L L

(3.43)

It will be noted that forces P1 and P2 are, in this case, statically equivalent to the applied actions F and M. The equations of motion for the system of Figure 3.11 are now obtained as  2     b  I0 I0 mb mab + − u¨ 1 k1 0 u1 FL − M 2 2 2 2 L L L L L   + = (3.44) a M I0 I0 mab ma2 0 k u ¨ u F + 2 2 2 − + 2 2 2 2 L L L

L

L

L

It is of interest to re-solve this problem with a new set of coordinates. As indicated in Figure 3.12a, these coordinates consist of vertical translation q1 at the mass center and rotation q2 about the mass center. With this set of coordinates, the mass matrix is easily shown to be given by   m 0 m= (3.45) 0 I0 To obtain the stiffness matrix, we first apply a unit displacement along coordinate 1, identify the internal elastic forces produced by such a displacement, and then calculate the external forces required to equilibrate the internal elastic forces. The relevant forces have all been indicated in Figure 3.12b. Using the conditions of equilibrium, we get k11 = k1 + k2 k21 = k2 b − k1 a

(3.46)

Next, we apply a unit displacement along coordinate 2 and calculate the external forces required to maintain the displacement. Referring to Figure 3.12c, these external forces

86

Dynamics of structures

Figure 3.12 Vibrating rigid bar; coordinates defined at the mass center.

are given by k12 = k2 b − k1 a

(3.47)

k22 = k1 a2 + k2 b2

The stiffness matrix is now assembled from the influence coefficients defined in Equations 3.46 and 3.47  K=

k1 + k2 k2 b − k 1 a

k2 b − k 1 a k1 a2 + k2 b2

 (3.48)

Since the applied forces act along the coordinates already defined, the force vector is obtained directly:  p=

F M

 (3.49)

The equations of motion can be assembled using Equations 3.45, 3.48 and 3.49 

m 0 0 I0



  k 1 + k2 q¨ 1 + q¨ 2 k2 b − k 1 a

k2 b − k 1 a k1 a2 + k2 b2



   q1 F = M q2

(3.50)

Formulation of the equations of motion: Multi-degree-of-freedom systems 87

A comparison of Equations 3.44 and 3.50 shows that while the stiffness matrix is diagonal in the first case, it is the mass matrix that is diagonal in the second case. A natural question that might arise is whether there is a set of coordinates in which both the mass matrix and the stiffness matrix are diagonal. There is indeed such a set of coordinates. Coordinates in the set are called normal coordinates and play a very important role in the field of structural dynamics. In later chapters we discuss such coordinate sets in considerable detail. Example 3.1 The uniform rigid rectangular slab of total mass m shown in Figure E3.1a is supported by three massless columns rigidly attached to the slab and fixed at the base. The columns have a flexural rigidity EI about each of the two principal axes, which are oriented so that they are parallel to the adjacent sides of the rectangle. Evaluate the mass and stiffness matrices for the system in the coordinates u1 , u2 , and u3 defined at the mass center.

Solution Because the columns are massless, they can be modeled by linear springs shown in Figure E3.1b. Each spring has a stiffness k = (12EI)/(L3 ). To obtain the first column of stiffness matrix, we impose a unit displacement in the direction of coordinate 1 and identify the internal spring forces that oppose the displacement. The external forces required to balance these spring forces, shown in Figure E3.1c, are now obtained from the conditions of equilibrium. k11 = 3k k21 = 0 k31

(a)

a a ka = 2k − k = 2 2 2

The second column of the stiffness matrix is obtained in a similar manner by imposing a unit displacement along coordinate direction 2. The resulting internal forces are shown in Figure E3.1d. The stiffness coefficients are given by k12 = 0 k22 = 3k k32 = −2k

(b) b kb b +k =− 2 2 2

The third column of stiffness matrix is obtained by imposing a unit rotation along coordinate 3. Referring to Figure E3.1e, we see that ka 2 kb =− 2
2  a 2  b 3  = 3k + 3k = k a2 + b 2 2 2 4

k13 = k23 k33

(c)

88

Dynamics of structures

Figure E3.1 Vibration of a rigid rectangular slab.

Formulation of the equations of motion: Multi-degree-of-freedom systems 89 The stiffness matrix is now assembled from Equations a, b, and c: 

3

 K = k 0 a 2

0

a 2

3

− b2

− b2

3 2 (a 4

   

(d)

+ b2 )

where k = 12EI/L3 . Since the selected coordinates are located at the mass center, the mass matrix is obtained directly and is given by  M=



m



m

(e)

I0 where I0 = m(a2 + b2 )/12.

3.3.4 Systems with distributed mass and distributed stiffness Systems having distributed mass as well as distributed stiffness are also referred to as continuous systems or distributed parameter systems. Theoretically, they have an infinite number of degrees of freedom and their motion is represented by partial differential equations. They can, however, be idealized as multi-degree-of-freedom systems either by mass lumping or by expressing their displaced shape as a superposition of a series of shape functions of the spatial coordinates each multiplied by its own generalized coordinate. Mass lumping was described briefly in Section 2.7.4. Equations of motion for lumped mass systems are obtained by the procedure described in Section 3.3.2. In this section we describe the procedure used to obtain the equations of motion for a distributed parameter system represented by a superposition of shape functions. The shape functions chosen to represent the displaced shape should be independent of each other; that is, it should not be possible to derive any one of them by a linear combination of one or more of the remaining. The functions should also, as a minimum, satisfy the geometric or essential boundary conditions of the system. Each shape function is multiplied by a generalized coordinate. The generalized coordinates then serve as the unknowns in the system, whose values must be determined by a solution of the equations of motion. The number of such coordinates is equal to the number of degrees of freedom in the system. As an illustration of the procedure, consider again the flexural vibrations of a beam supported at its ends. The vibration shape of the beam shown in Figure 3.13a is represented by a superposition of N shape functions ψ1 , ψ2 , . . . , ψN , so that u(x, t) = z1 ψ1 (x) + z2 ψ2 (x) + · · · + zN ψN (x)

(3.51)

The forces acting on an element of length dx are shown in Figure 3.13b. These forces include the flexural moments M, the inertia force m ¯ u¨ dx, and the externally applied force p¯ dx. We now obtain the virtual work equations for all admissible virtual displacements of the system. The admissible virtual displacements should satisfy the constraints in the system and should be independent. Since the shape functions ψi (x) are

90

Dynamics of structures

selected to satisfy the geometric or essential boundary conditions and are independent of each other, functions δzi ψi (x), i = 1, 2, . . . , N form a set of N admissible virtual displacement shapes. A virtual displacement applied to the beam causes the two ends of the infinitesimal element to rotate relative to each other by an angle δzi ψi (x) dx. Now in accordance with the elementary beam theory, moment M = EI(∂ 2 u/∂x2 ). The virtual work done by the elastic moments acting on the element is therefore given by d(δWei ) = δzi ψi (x)EI = δzi

N

∂ 2u dx ∂x2

zj EIψi (x)ψj (x) dx

(3.52)

j=1

The total virtual work done by the elastic moments is obtained by integrating Equation 3.52 over the length:  L N

zj EIψi (x)ψj (x) dx δWei = δzi 0

j=1

= δzi

N

zj kij

(3.53)

j=1

where



L

kij = 0

EIψi (x)ψj (x) dx

(3.54)

As explained in Section 2.6, the internal virtual work d(δWi ) is, in fact, the work done by the restoring elastic moments in the element. Since the restoring moments are opposite in sign to the moments shown in Figure 3.13b, work d(δWi ) is the negative of work d(δWei ) given by Equation 3.52. The virtual work done by the externally applied force is  L ¯ i (x) dx = pi δzi δWp = δzi pψ (3.55) 0

where



pi =

L

¯ i (x) dx pψ

(3.56)

0

In a similar manner, the virtual work of the inertia force is  L ∂ 2u ψi (x)m ¯ 2 dx δWI = −δzi ∂t 0  N L

z¨ j mψ ¯ i (x)ψj (x) dx = −δzi j=1

= −δzi

N

j=1

0

z¨ j mij

(3.57)

Formulation of the equations of motion: Multi-degree-of-freedom systems 91

Figure 3.13 (a) Flexural vibrations of a beam; (b) forces acting on an element; (c) internal damping resistance; (d) axial force effect.

where



mij =

L

mψ ¯ i (x)ψj (x) dx

(3.58)

0

Equating the sum of virtual works δWi , δWp , and δWI to zero, we get the following virtual work equation   N N



−δzi  z¨ j mij + zj kij  + δzi pi = 0 j=1

(3.59)

j=1

Since δzi is arbitrary, it can be canceled from Equation 3.59. Corresponding to the N independent virtual displacements δzi ψi (x), i = 1, 2, . . . , N, there are N virtual work equations of the form of Equation 3.59. Together they can be expressed in matrix notations as M¨z + Kz = p

(3.60)

where z is a vector of N generalized coordinates, M is a mass matrix whose elements are given by Equation 3.58, K is a stiffness matrix whose elements are defined by Equation 3.54, and p is the vector of generalized forces given by Equation 3.56. By referring to Equation 3.58 for mass influence coefficients, it is seen that mij = mji . In a similar manner, Equation 3.54 shows that kij = kji . Thus both mass matrix M and stiffness matrix K are symmetric.

92

Dynamics of structures

In formulating the virtual work done by internal spring forces, we have neglected the work done by the internal shear forces on virtual shear deformations. For normal proportions of beams, that is, when the cross-sectional dimensions are small in relation to the span length, the work done by shear forces is negligible in comparison to that done by the flexural moments. Equation 3.60 does not include damping forces present in the system. Such forces may either be external or internal. External damping can be provided by distributed viscous damping forces, as indicated in Figure 2.16d. Denoting the distributed viscous damping coefficient by c¯ , the virtual work done by external damping forces is obtained as 

L

δWDE = −δzi

c¯ ψi (x)

0

= −δzi

N  L

j=1

∂u dx ∂t

z˙ j c¯ ψi (x)ψj (x) dx

(3.61)

0

Internal damping forces resist deformations within the element and their magnitudes depend on the strain rate. Thus if the strain rate is ∂ε/∂t, the damping resistance can be represented by a stress σD which is proportional to the strain rate, the constant of proportionality being a damping constant cs . Thus ∂ε ∂t

σD = cs

(3.62)

Referring to Figure 3.13c and using elementary beam theory, which assumes that plane sections remain plane under bending, we get the following kinematic relationship: ε = κy =

∂ 2u y ∂x2

(3.63)

where κ is the curvature of the beam. From Equations 3.62 and 3.63, ∂ 3u y ∂x2 ∂t

σD = cs

(3.64)

The resisting moment due to internal damping is given by  MD =

σD y dA A

 =

cs y 2 A

= cs I

∂ 3u dA ∂x2 ∂t

∂ 3u ∂x2 ∂t

(3.65)

Formulation of the equations of motion: Multi-degree-of-freedom systems 93

Finally, recognizing that under a virtual displacement the ends of the element undergo a relative rotation of δzi ψi (x) dx, the virtual work done by the internal damping moment works out to 

L

δWDI = −δzi 0

= −δzi

cs Iψi (x)

N  L

j=1

0

∂ 3u dx ∂x2 ∂t

cs I z˙ j ψi (x)ψj (x) dx

(3.66)

Combining Equations 3.61 and 3.66, the total work done by the damping forces can be expressed as δWD = −δzi

N



N



L

c¯ ψi (x)ψj (x) dx +

0

j=1

= −δzi

L

z˙ j

0



cs Iψi (x)ψj (x) dx

z˙ j cij

(3.67)

j=1

where  cij =

L



L

c¯ ψi (x)ψj (x) dx +

0

0

cs Iψi (x)ψj (x) dx

(3.68)

When damping is included, the equations of motion (Eq. 3.60) are revised to M¨z + C˙z + Kz = p

(3.69)

where C is a damping matrix whose elements are defined by Equation 3.68. In practice, it is difficult to define damping constants c¯ and cs on the basis of physical characteristics of the system. Alternative methods are therefore used to include damping resistance in the model. The suggested methods lead to a response characteristic that correlates well with experimental or observed behavior. The methods of defining the damping forces or the damping matrix C are considered in the subsequent chapters. When concentrated applied forces, masses, springs, or dampers are present in the system, they are handled in a manner similar to that described in Section 2.7.4. If axial forces are present, the equations of motion (Eq. 3.69) will need modification. As discussed earlier, the effect of axial forces can be allowed for by deriving a geometric stiffness matrix which must be deducted from the elastic stiffness matrix of the beam. The derivation of the geometric stiffness matrix follows a procedure very similar to that used in the single-degree-of-freedom representation of the beam. Thus referring to Figure 3.13d, the virtual displacement δzi ψi (x) will cause the two axial forces S(x) to move closer to each other by a distance δzi (∂u/∂x)ψi (x) dx, in which

94

Dynamics of structures

∂u/∂x is obtained from Equation 3.51. The virtual work done by the axial forces is therefore given by d(δWS ) = δzi

N

zj S(x)ψj (x)ψi (x) dx

(3.70)

j=1

The total virtual work is obtained by integrating Equation 3.70 over the length: δWS = δzi

N 

j=1

= δzi

N

L 0

zj S(x)ψi (x)ψj (x) dx

zj kG ij

(3.71)

j=1

where  kG ij = 0

L

S(x)ψi (x)ψj (x) dx

(3.72)

It should be noted that δWS is positive while δWi was negative. The equation of motion, including the axial force effect, becomes M¨z + C˙z + (K − KG )z = p

(3.73)

where KG is the geometric stiffness matrix whose elements are defined by Equation 3.72. The procedure described in the foregoing paragraphs in which the displaced shape is expressed as a superposition of a series of appropriately selected shape functions is also known as the Ritz method. The shape functions selected to represent the displaced configuration of the system are called the Ritz shapes. While conceptually elegant, the Ritz method poses several difficulties in its practical application. These difficulties are outlined below. 1

2

It is apparent that the success of the Ritz method depends on the selection of shape functions. This is, in general, a difficult task. As a minimum, the shape functions should satisfy the essential boundary conditions of the problem. In addition, if the highest-order differential appearing in the virtual work equation is of order m, the shape functions should be m times differentiable. In other words, shape functions should satisfy C m−1 continuity; that is, their (m−1)th differential should be continuous. The choice of shape functions should thus be guided by the nature of the problem and the boundary conditions, and it is not always apparent what shape functions would be appropriate in a particular case. The shape functions should span the entire domain of the system, yet the displacement being represented may vary in widely different manner in different regions of the domain. As an example, even in the simple case of the flexural vibrations

Formulation of the equations of motion: Multi-degree-of-freedom systems 95

3 4

5

6

of a beam, if the moments of inertia in different sections of the length are significantly different, shape functions that are appropriate for one region may not be appropriate for another. A flat plate with edge beams is another example where displacements may vary in widely differing manners. It is not always clear how the Ritz method should be refined, or what additional Ritz shapes should be included to improve the accuracy of the solution. The virtual work expressions used in setting up the equations of motion involve integration of the shape function derivatives and their products. Unless the shape functions are simple mathematical functions, such integration may not be straightforward. The property matrices obtained in a Ritz formulation are fully populated. As a result, when the number of Ritz vectors is large, the solution becomes computationally inefficient. The generalized coordinates used as the unknown weights on the shape functions do not always have a physical meaning. It is therefore difficult to interpret the behavior of the system being analyzed from the generalized coordinate values obtained in the analysis.

The difficulties involved in the Ritz function approach are largely overcome in a finite element formulation, which is discussed in Chapter 18. Example 3.2 The vibration shape of the simply supported uniform beam of length L, flexural rigidity EI, and mass m ¯ per unit length shown in Figure E3.2 is approximated by u(x, t) = z1 (t)ψ1 (x) + z2 (t)ψ2 (x) + z3 (t)ψ3 (x)

(a)

where πx L 2πx ψ2 (x) = sin L 3πx ψ3 (x) = sin L

ψ1 (x) = sin

(b)

Figure E3.2 Flexural vibrations of a uniform beam.

96

Dynamics of structures

Obtain the equations of motion when the beam is vibrating under the action of a uniformly ¯ distributed load p(t).

Solution We have ψm (x) = sin ψm (x)

mπx L

(c)

mπx m2 π 2 = − 2 sin L L

Also, 

L

sin 0

nπ x mπ x sin dx = L L



0 L 2

m = n m=n

(d)

The elements of stiffness matrix are obtained from Equation 3.54. Substituting from Equations c and d in Equation 3.54, we get kmn = 0 kmm

m = n

m4 π 4 EI = 2L3

(e)

The complete stiffness matrix is given by  1 π 4 EI  0 K= 2L3 0

 0 0 16 0  0 81

(f)

The mass matrix is obtained by using Equation 3.58 and can be shown to be  1 mL ¯  0 M= 2 0

0 1 0

 0 0 1

(g)

The elements of the load vector obtained from Equation 3.56 are pm = p¯



L

sin 0

mπ x dx L

The load vector reduces to   1 ¯ 2pL   p= 0 π 1

(h)

(i)

3

The equation of motion is M¨z + Kz = p where M, K, and p are defined by Equations g, f, and i, respectively.

(j)

Formulation of the equations of motion: Multi-degree-of-freedom systems 97 It is a coincidence that in this case both the mass matrix and the stiffness matrix are diagonal and Equation j is therefore uncoupled; that is, it is equivalent to three independent equations, each of which can be solved separately. In fact, uncoupling of the equations of motion has been made possible by the selection of shape functions that are orthogonal to each other. Coordinates that uncouple the equations of motion are called normal coordinates. Such coordinates exist for every dynamic system, whether discrete or continuous. It is obvious that a formulation of the equations of motion in terms of normal coordinates will make the solution of such equations much simpler. However, it is not always easy to recognize the normal coordinates of a system. In later chapters, we describe in considerable detail procedures for obtaining the normal coordinates.

Example 3.3 The uniform beam in Figure E3.3a has four independent displacement degrees of freedom, as indicated in the figure. Two of these are the rotations at the ends of the beam, while the other two are vertical displacements at the same locations. By applying appropriate external forces along the four degrees of freedom, it is possible to obtain a displaced configuration of the beam in which there is a unit displacement along one of the four degrees of freedom while the displacements along the remaining three degrees of freedom are zero. Corresponding to the

Figure E3.3 Flexural displacements of a prismatic beam.

98

Dynamics of structures

four degrees of freedom, there are four such independent displaced shapes. By using elementary beam theory, obtain these four shapes. Then, by expressing the displacement of the beam as a superposition of the four basic shapes, obtain the stiffness matrix of the beam.

Solution According to elementary beam theory, the deflections of a uniform beam subjected to loads only at its ends are governed by the following differential equation: EI

d4 ψ =0 dx4

(a)

Equation a has a solution of the form ψ = Ax3 + Bx2 + Cx + D

(b)

where A, B, C, and D are constants to be determined from the boundary conditions. For the configuration shown in Figure E3.3b, these boundary conditions are ψ =0

and

∂ψ =1 ∂x

x=L ψ =0

and

∂ψ =0 ∂x

x=0

(c)

When constants A, B, C, and D are obtained by using Equation c, displacement ψ is given by  x 2 ψ1 (x) = x 1 − L

(d)

Displaced shapes shown in Figure E3.3c, d, and e are obtained in a similar manner and are given by  x 2

ψ2 (x) = 1 − 3

 x 3

+2

L L  x2  x −1 ψ3 (x) = L L  x 2  x 3 −2 ψ4 (x) = 3 L L

(e)

The displaced shape of the beam can be expressed as a superposition of the shape functions given by Equations d and e, so that u = z1 ψ1 (x) + z2 ψ2 (x) + z3 ψ3 (x) + z4 ψ4 (x)

(f)

The elements of the stiffness matrix are now obtained by using Equation 3.54. The resulting stiffness matrix is given by 

4EI L

6EI L2

2EI L

6EI L2

12EI L3

6EI L2

2EI L

6EI L2

4EI L

− 6EI L2

− 12EI L3

− 6EI L2

   K=  

− 6EI L2



  − 12EI L3    − 6EI L2  12EI L3

(g)

Formulation of the equations of motion: Multi-degree-of-freedom systems 99 We now derive the external forces that must be applied along the four degrees of freedom defined earlier to maintain a specified displaced configuration, say the one in Figure E3.3b. These forces can be obtained either by direct application of the beam theory or by the virtual work method. Thus if the beam in Figure E3.3b is given a virtual displacement of δz2 ψ2 (x), the virtual work done by the moments acting on internal elements is  δWei = δz2

L

0

EIψ1 (x)ψ2 (x) dx

(h)

At the same time, of the four external forces shown in Figure E3.3b, only the force f21 does any work, the virtual displacements along the other three forces being zero. This can be verified from Figure E3.3c, which also represents the virtual displacement shape. The virtual work done by the external forces becomes δWe = δz2 f21

(i)

Using the virtual work equation yields δz2 f21 = δWei

(j)

or  f21 =

L

0

EIψ1 (x)ψ2 (x) dx = k21

(k)

Other forces shown in Figure E3.3b can be derived similarly and are seen to be equal to the elements on the first column of the stiffness matrix. Elements on other columns of the stiffness matrix are interpreted in a similar manner. In conclusion, it can be stated that when functions given by Equations d and e are used as the shape functions, the stiffness matrix derived by the procedure of Equation 3.54 is, in fact, the stiffness matrix corresponding to the four degrees of freedom defined in Figure E3.3a, and the weighting factors z1 through z4 in Equation f represent displacements along these degrees of freedom.

Example 3.4 The uniform bar shown in Figure E3.4 has an area A and mass per unit length m. ¯ The bar is ¯ vibrating in the axial directions under the action of a distributed axial force p(x). Express the vibration shape of the bar as a superposition of appropriate shape functions, selecting for the generalized coordinates the longitudinal displacements at the two ends. Then obtain the stiffness and mass matrices and the force vector corresponding to the two generalized coordinates defined above.

Solution The vibration shape of the bar is represented by a superposition of two shape functions, ψ1 and ψ2 , weighted by the generalized coordinates u1 and u2 shown in Figure E3.4a. u(x, t) = u1 (t)ψ1 (x) + u2 (t)ψ2 (x)

(a)

100

Dynamics of structures

Figure E3.4 Axial vibrations of a uniform bar: (a) elevation of the bar showing displacement coordinates; (b) forces acting on the bar element.

We select the following shape functions: ψ1 = 1 − ψ2 =

x L

x L

(b)

The shape functions have been selected so that at x = 0, ψ1 = 1 and ψ2 = 0 while at x = L, ψ1 = 0 and ψ2 = 1. With the selected shape functions, Equation a gives a displacement u = u1 at x = 0 and a displacement u = u2 at x = L. We now apply appropriate virtual displacements to the bar and obtain an equation of virtual work for each applied virtual displacement. A set of admissible virtual displacements of the system consists of δui ψi (x), i = 1, 2. A virtual work equation is obtained for each displacement in the set. The forces acting on an element of length dx are shown in Figure E3.4b. They consist of the elastic axial forces P and the inertia forces m ¯ u¨ dx. A virtual displacement applied to the bar causes the two ends of the infinitesimal element to move relative to each other in an axial direction by a distance δui ψi dx. Since the axial force P is given by P = EA(∂u/∂x), the virtual work done by the elastic forces acting on the element is given by d(δWei ) = δui ψi (x)EA = δui

2

∂u dx ∂x

uj EAψi (x)ψj (x) dx

(c)

j=1

The total virtual work done by the elastic forces is obtained by integrating Equation c over the length:

δWei = δui

2

j=1

= δui

2

j=1



L

uj 0

uj kij

EAψi (x)ψj (x) dx

(d)

Formulation of the equations of motion: Multi-degree-of-freedom systems 101 where the stiffness influence coefficients kij are given by 

L

kij = 0

EAψi (x)ψj (x) dx

(e)

Substitution for ψi and ψj in Equation e gives the following values for the stiffness influence coefficients: 
1 2 EA EA − dx = L L 0 
 L EA 1 1 dx = − = k21 = EA − L L L 0
2  L 1 EA EA dx = = L L 0 

k11 = k12 k22

L

(f)

Assembly of the stiffness influence coefficients gives the following stiffness matrix: K=

 EA 1 L −1

−1 1

 (g)

The virtual work of the inertia forces is obtained from 

L

δWI = −δui

mψ ¯ i (x)

0

= −δui

2 

j=1

= −δui

2

L

∂ 2u dx ∂t 2

m ¯ u¨ j ψi (x)ψj (x) dx

0

u¨ j mij

(h)

j=1

where the mass influence coefficients mij are given by 

L

mij =

mψ ¯ i (x)ψj (x) dx

(i)

0

Substitution for ψi and ψj in Equation i leads to the following value for the mass influence coefficients: 

 x 2 L m ¯ 1− dx = m ¯ L 3 0  L   L x x dx = m ¯ = m21 = m ¯ 1− L L 6 0  L  2 x L = m ¯ dx = m ¯ L 3 0

m11 = m12 m22

L

(j)

102

Dynamics of structures

Assembly of the mass influence coefficients gives the following mass matrix: 1 M = mL ¯

3

1 6

1 6

1 3

 (k)

The virtual work done by the externally applied force is 

L

δWp = δui

¯ p(x, t)ψi (x) dx

0

= pi δui

(l)

where  pi =

L

¯ p(x, t)ψi (x) dx

(m)

0

¯ For a uniformly distributed load p(t), Equation m yields ¯ p1 = p(t)



L



1−

0

L ¯ = p(t) 2  ¯ p2 = p(t)

L

0

x dx L

(n) x dx L

L ¯ = p(t) 2 The generalized force vector becomes L p = p¯ 2

 1 1

(o)

3.4 TRANSFORMATION OF COORDINATES Methods of formulating the equations of motion in a selected set of coordinates, either physical or generalized, were discussed in the preceding sections. Often, it may be necessary or convenient for the analysis of response to express the equations in a set of coordinates that are different from the ones in which the equations were initially formulated. Such a transformation can easily be achieved by using the principle of virtual work. Suppose, for instance, that the equations have originally been formulated in a set of N coordinates denoted by u, and let the corresponding mass matrix, stiffness matrix, and applied force vector be denoted by M, K, and p, respectively. Now let it be required to transform the equations of motion to a set of N independent coordinates q which are related to the set u by the equation u = Tq

(3.74)

Formulation of the equations of motion: Multi-degree-of-freedom systems 103

where T is a transformation matrix, which is square. Let the transformed mass and stiffness matrices and the applied force vector be denoted by M, K, and p, respectively. The spring forces in u set of coordinates are fS = Ku

(3.75)

while those in q set of coordinates are f˜S = Kq

(3.76)

Now let the system be subjected to a set of virtual displacement δq. The corresponding displacements in the u set are δu, where δu = T δq

(3.77)

If the forces f˜S and fS are equivalent, the virtual work done by each set should be the same. Thus δqT Kq = δuT Ku

(3.78)

Substituting for u and δu from Equations 3.74 and 3.77, respectively, we get δqT Kq = δqT TT KTq

(3.79)

Since Equation 3.79 should hold for arbitrary q and δq, we get K = TT KT

(3.80)

In a similar manner, considering the virtual work done by inertia forces, we can show that M = TT MT

(3.81)

Consideration of virtual work done by the applied forces gives p = TT p

(3.82)

If damping is present in the system, the damping matrix can be transformed in a similar manner, giving C = TT CT

(3.83)

Example 3.5 In Section 3.3.3, equations of motion were obtained for a spring-supported rigid bar in two different sets of coordinates, shown in Figures 3.4a and 3.12a, respectively. Starting from the formulation in the u set of coordinates (Fig. 3.4a) and a transformation relationship, obtain the equations of motion in the q set of coordinates shown in Figure 3.12a.

104

Dynamics of structures

Solution The transformation from the q set to the u set can be expressed as u = Tq

(a)

where T is a transformation matrix given by 

1 T= 1

−a b

 (b)

The transformed stiffness matrix K is obtained from Equation 3.80: K = TT KT     k1 0 1 −a 1 1 = 1 b −a b 0 k2   k1 + k2 bk2 − ak1 = 2 2 bk2 − ak1 a k1 + b k2

(c)

which is the same as that obtained directly in Equation 3.48. The transformed mass matrix is obtained from Equation 3.81: M = TT MT  mb2  + 1 1 l2 = mab −a b − l2   m = I0

Io l2 Io l2

mab l2 ma2 l2

− +

Io l2 Io l2



1 1

−a b



(d)

and is the same as that given by Equation 3.45. The transformed applied force vector is given by p = TT p  Fb  − 1 1 l = Fa −a b + l   F = M

M l M l



(e)

which is equal to that obtained directly in Equation 3.49.

Example 3.6 The uniform bar shown in Figure E3.6 undergoes axial vibrations. Obtain the property matrices and the force vector of the bar in terms of generalized coordinates q1 through q4 shown in the figure. The longitudinal axis of the bar makes an angle α with respect to the coordinate direction q1 .

Solution The stiffness and mass matrices as well as the force vector for the bar were obtained in Example 3.4 corresponding to the coordinates u1 and u3 shown in Figure E3.6. We can obtain the

Formulation of the equations of motion: Multi-degree-of-freedom systems 105

Figure E3.6 Transformation of coordinates for a bar in axial vibrations.

structural property matrices and the force vector in the generalized coordinate set q by means of a simple transformation. Before carrying out the transformation, we need to expand the property matrices and the force vector obtained earlier to an order 4 to correspond to the four coordinates u1 through u4 . This is easily achieved by adding appropriate number of rows and columns of zero. The revised stiffness matrix is given by 

EA L

 0  Ku =  EA − L 0



0

− EA L

0

0

0

EA L

0   0

0

0

0

0

(a)

where we have added a row and a column of zeros in each of positions 2 and 4. Physically, this represents the fact that movements of the bar along coordinate directions 2 and 4 represent rigid-body motions which induce no stresses. The mass matrix is expanded in the same manner as the stiffness matrix. However, since motion in the direction perpendicular to the axis of the bar will also attract inertia forces, the mass terms corresponding to a direction along the axis are repeated in the direction perpendicular to the axis and we get: 1 3

0

1 6

0 1 6



0  Mu = mL ¯ 1 6

1 3

0

0

1 3

   0

0

1 6

0

1 3

The force vector becomes   1 0 ¯ p(t)L   2 1 0

(b)

(c)

106

Dynamics of structures

The transformation between the q and u set of coordinates is easily shown to be 

cos α  − sin α  u= 0 0

sin α cos α 0 0

0 0 cos α − sin α

 0 0  q sin α  cos α

(d)

or u = Tq

(e)

where T represents the transformation matrix. The transformed stiffness matrix is given by Kq = TT Ku T

(f)



c2  EA  cs = L  −c2 −cs

cs s2 −cs −s2

−c2 −cs c2 cs



−cs −s2   cs  2 s

(g)

where c ≡ cos α and s ≡ sin α. The transformed mass matrix is obtained from Mq = TT Mu T 1 3

(h) 0

1 6

0 1 6



0  = mL ¯ 1 6

1 3

0

0

1 3

   0

0

1 6

0

1 3

(i)

It is noted that the transformation leaves the mass matrix unaltered. The transformed force vector is given by pq = TT pu   c ¯ s pL   = 2 c s

3.5

(j)

(k)

STATIC CONDENSATION OF STIFFNESS MATRIX

In general, the kinetic energy component corresponding to the rotational degrees of freedom is negligible in comparison to that corresponding to the translational degrees of freedom. Consequently, the inertial moments along the rotational degrees of freedom can be ignored and the rotational degrees of freedom eliminated from the mass matrix. Further, if there are no other loads acting in the direction of the rotational degrees of freedom, the spring forces along these degrees of freedom should also be zero. This condition allows us to eliminate the rotational coordinates from the stiffness matrix

Formulation of the equations of motion: Multi-degree-of-freedom systems 107

as well, permitting a significant reduction in the size of the problem. The reduced problem usually is much easier and less expensive to solve. For example, if the displacement vector is partitioned so that ut represents the translational degrees of freedom and uθ represents the rotational degrees of freedom, and the stiffness and mass matrices are also partitioned accordingly, the equations of motion can be written as 

Mtt 0

0 0



u¨ t u¨ θ





Ktt + Kθt

Ktθ Kθθ



ut uθ





pt = 0

 (3.84)

where it is assumed that the rotational degrees of freedom have not been included in the mass formulation and that no applied loads act along such degrees of freedom. Expansion of Equation 3.84 gives Mtt u¨ t + Ktt ut + Ktθ uθ = pt

(3.85)

Kθt ut + Kθθ uθ = 0

(3.86)

Equation 3.86 leads to −1 Kθt ut uθ = −Kθθ

(3.87)

Substitution of Equation 3.87 in Equation 3.85 gives Mtt u¨ t + K∗ ut = pt

(3.88a)

where −1 Kθt K∗ = Ktt − Ktθ Kθθ

(3.88b)

Matrix K∗ is called the condensed stiffness matrix. The reduced problem of Equation 3.88a is generally significantly smaller in size than the original problem represented by Equation 3.84. Example 3.7 The stiffness matrix for a simply supported beam of Figure 3.10a is initially formulated in the four degrees of freedom shown in Figure E3.7a. These include two rotational degrees of freedom indicated in the figure. The mass matrix (Eq. 3.27) and the load vector (Eqs. 3.35 and 3.36), however, include only the translational degrees of freedom. Obtain first the stiffness matrix corresponding to the four degrees of freedom shown in Figure E3.7a, and then eliminate the two rotational degrees of freedom to obtain a reduced stiffness matrix.

108

Dynamics of structures

Figure E3.7 Vibrations of a simply supported beam: matrix condensation.

Solution Stiffness matrices of the three sections of the beam: AB, BC, and CD corresponding to the local coordinates shown in Figure E3.7b are obtained by standard methods of structural analysis and are given by

K

AB

= EI

(1)

(3)

81 L3 − L272

− L272

! (1) (3)

 (1)

K

BC

(3)

(2)

324 L3

54 L2

− 224 L3

54 L2

12 L

− L542

6 L

− L542

324 L3

− L542

6 L

− L542

12 L

L2

(2) (4)

(a)

9 L

(1)  54 (3)   L2 = EI  (2) − 324  L3 (4) 54

KCD = EI

"

(2) ! 81

(4) " 27

L3

L2

27 L2

9 L

(4)       

(b)

(c)

In Equations a, b, and c, the positions that individual elements of the given matrices will occupy after assembly in the global stiffness matrix have been indicated by specifying the global matrix row and column numbers that correspond to the rows and columns of the component matrices. This information is easily obtained by noting the correspondence between the element

Formulation of the equations of motion: Multi-degree-of-freedom systems 109 and global degrees of freedom. For example, in Equation b, element k23 = −54/L2 will be placed in row 3 and column 2 of the global matrix. The 4 × 4 global stiffness matrix is now obtained by direct stiffness assembly, giving 

405 L3

 324 − 3  L K = EI   272  L 54 L2

− 324 L3

27 L2

405 L3

− L542

− L542

21 L

− L272

6 L

54 L2



 − L272    6  L 

(d)

21 L

Partitioning the stiffness matrix Equation d along translational and rotational degrees of freedom, we get Ktt = EI

− 324 L3

Ktθ = KθTt = EI 21 Kθθ = EI

− 324 L3

405 L3

405 L3 27 L2

− L542  6

L

L

6 L

21 L



54 L2



− L272

Substitution into Equation 3.88b gives the condensed stiffness matrix  4 − 72 324EI ∗ K = 5L3 − 72 4

(e)

(f)

The stiffness matrix in Equation f is the same as that obtained in Equation 3.31 by first forming the flexibility matrix corresponding to the two translational degrees of freedom and then taking its inverse. In practice, the latter approach in which the stiffness matrix is obtained as an inverse of the flexibility matrix is usually simpler and computationally more efficient than direct static condensation using Equation 3.88b, particularly when the size of the reduced matrix is significantly smaller than that of the original matrix.

3.6 APPLICATION OF RITZ METHOD TO DISCRETE SYSTEMS In Section 3.3.4 we described the application of Ritz method to the representation of a distributed parameter system having an infinite number of degrees of freedom by a system having a finite number of degrees of freedom. The Ritz procedure can also be used to represent the displaced shape of a discrete multi-degree-of-freedom system by the superposition of a few appropriately selected shape functions. Analogous to Equation 3.51, we have u=

M

zi ψ i

(3.89)

i=1

where u is the displacement shape vector of the original N-degree-of-freedom system, ψi are shape function vectors, and zi are generalized coordinates. We note that instead

110

Dynamics of structures

of being continuous function of spatial coordinates, ψi ’s are now vectors of size N. In matrix notation, Equation 3.89 can be expressed as u = z

(3.90)

where  is an N by M matrix of shape function vectors and z is a vector of size M. When the number of shape vectors, M, is equal to N, Equation 3.90 is entirely equivalent to the standard transformation equation (Eq. 3.74). Vectors  are in this case said to span the entire N-dimensional space and the solution of transformed equations will lead to the same response as the original equations. As stated earlier, transformation of this type may be useful if the transformed equations are easier to solve. In general, however, the usefulness of the Ritz procedure is most evident when only a few shape function vectors are adequate to represent the response. In such a case, M is much smaller than N. Transformation equations 3.80 through 3.83 are still applicable, except that the transformation matrix  is rectangular of size N by M and the transformed property matrices are of size M by M. Thus M = T M

(3.91)

K = T K

(3.92)

C =  C

(3.93)

p= p

(3.94)

T T

and the transformed equations of motion are M¨z + C˙z + Kz = p

(3.95)

Equations 3.95 are of order M and because M  N, the reduced problem is of a significantly smaller size than the original. Example 3.8 The four-story shear-type building frame shown in Figure E3.8 has a concentrated mass m at each story level. The total column stiffness for each story is k. The frame is subjected to a lateral dynamic force of p0 sin t applied at story level 4. Using the following two shape function vectors, obtain the equations of motion of a reduced system of order 2.  0.250  0.500   ψ1 =   0.750  1.000

 −0.643  −0.929   ψ2 =   −0.500  1.000





Solution Corresponding to the four degrees of freedom shown in Figure E3.8, the mass and stiffness matrices are given by the following equations: 

1 0 M = m 0 0

0 1 0 0

0 0 1 0

 0 0  0 1

(a)

Formulation of the equations of motion: Multi-degree-of-freedom systems 111

Figure E3.8 Lateral vibrations of a shear frame. 

2  −1 K = k  0 0

 −1 0 0 2 −1 0  −1 2 −1  0 −1 1

(b)

The applied force vector is given by  pT = p0 sin t 0

0

0

1



(c)

The transformation matrix  is obtained from the given values of ψ1 and ψ2 , so that 

0.250  0.500 =  0.750 1.000

 −0.643 −0.929   −0.500  1.000

(d)

The transformed mass and stiffness matrices are now obtained from Equations 3.91 and 3.92, respectively:  M= 

1.875 0 0 2.526

0.250 K= 0.250

0.250 2.929

 (e)

 (f)

The transformed applied force vector obtained from Equation 3.116 becomes p=

  1 p sin t 1 0

(g)

It may be of interest to note that the Ritz vectors used in this case leave the mass matrix diagonal but do not diagonalize the stiffness matrix. This is because the procedure that was used

112

Dynamics of structures

to obtain the Ritz vectors shown here orthogonalizes the vectors produced with respect to the mass matrix. We discuss this procedure in a later chapter.

SELECTED READINGS Clough, R.W., “Analysis of Structural Vibrations and Dynamic Response’’, Recent Advances in Matrix Methods of Structural Analysis and Design, (Eds. R. H. Gallagher, Y. Yamada, and J.T. Oden), University of Alabama Press, Huntsville, Ala, 1971, pp. 441–486. Clough, R.W., and Penzien, J., Dynamics of Structures, 2nd Edition, McGraw-Hill, New York, 1993. Meirovitch, L., Analytical Methods in Vibrations, Macmillan, London, 1967. Warburton, G.B., The Dynamical Behavior of Structures, 2nd Edition, Pergamon Press, Oxford, U.K., 1976. Warburton, G.B., “Response Using the Rayleigh Ritz Method’’, Earthquake Engineering and Structural Dynamics, Vol. 7, 1979, pp. 327–334.

PROBLEMS 3.1

An automobile and its suspension system are idealized as shown in Figure P3.1. The vehicle body along with its payload is considered to be a rigid body with total weight W, and radius of gyration r for pitching motion about an axis perpendicular to the paper and passing through the centroid. The suspension system supporting each axle is represented by a spring of stiffness kS and a viscous damper with damping coefficient cS . Weights w applied to the top of front and rear tires represent the weight of chassis and axles. The stiffness of a pair of tires is kT . The vehicle undergoes vibrations involving bounce and pitch. Identify the degrees of freedom of the system and obtain the equations of motion.

Figure P3.1

Formulation of the equations of motion: Multi-degree-of-freedom systems 113 3.2

The rocking behavior of certain types of structures under the action of earthquake motion can be studied by analyzing the model shown in Figure P3.2. The structure is represented by a block of mass m and mass moment of inertia about its centroid I0 . The foundation soil is modeled by a pair of springs each of stiffness k and a pair of viscous dampers of damping constants c. Assuming that the structure does not slide laterally with respect to the foundation or lift off its foundation, obtain the equations of motion for a ground acceleration v¨ g in the vertical direction.

Figure P3.2 3.3

A platform consisting of precast concrete planks is supported by three steel frames as shown in Figure P3.3. Frames A and B, which are identical, run parallel to the X direction. Frame C is parallel to the Y direction. The lateral stiffness of each of frames A and B is k1 while that of frame C is k2 . Obtain the equations of motion for free vibrations

Figure P3.3

114

Dynamics of structures of the deck in terms of the three coordinates defined at the center of mass as shown in Figure P3.3.

3.4

Figure P3.4 shows the plan views of a small two story building. The structure consists of three frames A, B and C oriented as shown. The stiffness matrices of the frames, defined along the coordinate directions indicated, are given in Figure P3.4. The floor weight is uniformly distributed and has a magnitude of 100 lb/ft2 . The building is subjected to a time varying uniform pressure p0 sin t lb/ft2 on its face XY. Obtain the equations of motion governing the response of the building in terms of coordinates θ1 , θ2 , θ3 and θ4 .

Figure P3.4 3.5

An industrial crane of span 80 ft runs along two crane girders which are supported on a span of 20 ft (Fig. 3.5). The mass of the crane girders and rails may be neglected. The mass of the crane itself may be assumed to be lumped at the center of its span and at the two ends as shown. Obtain the equations of free vibrations of the crane and the hoist trolley, assuming that the trolley is at the center of the span and does not carry any load and that the crane is in the middle of the crane girder. The following data is supplied.

Formulation of the equations of motion: Multi-degree-of-freedom systems 115

Figure P3.5 Moment of inertia of crane girder, I1 = 4090 in.4 Moment of inertia of crane structure, I2 = 36000 in.4 Total weight of the crane = 55000 lb Weight of hoist trolley = 7500 lb E = 30000 ksi 3.6

Obtain the equations of motion for the free vibrations of the system shown in Figure P3.6.

Figure P3.6 3.7

The double pendulum shown in Figure P3.7 consists of two simple pendulums connected by a spring of stiffness k attached at a distance a from the point of suspension. The length of each pendulum is l and the pendulum masses are m1 and m2 , respectively. Obtain the equations of motion for small vibrations of the double pendulum.

Figure P3.7

116 3.8

Dynamics of structures Obtain the equations of motion for the frame in Figure P3.8, which is vibrating under the action of a vertical load P(t) acting as shown.

Figure P3.8 3.9

The chimney referred to in Problem 2.9 is vibrating under the action of a lateral ground motion having an acceleration u¨ g (t). The vibrations of the chimney can be represented by superposition of the following two mode shapes: πx 2L 3πx ψ2 = 1 − cos 2L

ψ1 = 1 − cos

Obtain the equations of motion: (a) ignoring the gravity load effect; (b) taking gravity load effect into account. Assume that damping is negligible. 3.10

A uniform slab is supported by three identical frames as shown in Figure P3.10. Derive the stiffness matrix of an individual frame with reference to the three degrees of freedom

Figure P3.10

Formulation of the equations of motion: Multi-degree-of-freedom systems 117 indicated in the figure. Assume that the beam as well as the columns are axially rigid. Obtain the lateral stiffens of a frame by static condensation of the stiffness matrix to eliminate degrees of freedom u1 and u2 . Then obtain the equations of motion for free vibrations of the slab-frame system in terms of the three degrees-of-freedom defined about the mass center. The weight of the slab including the superimposed dead load is 100 lb/ft2 . 3.11

A vehicle traveling across a bridge deck is represented by an unsprung mass mt connected to a sprung mass mv through a spring of stiffness k and a damper of coefficient c (Fig. P3.11). The bridge deck has a span L, a uniform mass m ¯ per unit length and flexural rigidity EI. The vehicle is traveling across the deck at a constant speed of v. If the vibration shape of the deck is ψ(x) = sin(πx/L), obtain the equations of motion for the system assuming that the vehicle damping c and unsprung mass mt are negligible.

Figure P3.11 3.12

Repeat Problem 3.11 taking vehicle damping c and unsprung mass mt into account.

3.13

Obtain the equations of motion for free vibrations of the system of Problem 2.4 assuming that the wheel is braked to its axle.

3.14

Derive the equations of free vibration for the concrete plane frame ABC shown in Figure P3.14. The moment of inertia of leg AB may be assumed to vary linearly from

Figure P3.14

118

Dynamics of structures 10 × 106 mm4 at A to twice that value at B. The mass of AB also varies linearly from 69 kg/m at A to 138 kg/m at B. Beam BC has a uniform cross-section with a moment of inertia 20 × 106 mm4 and a mass of 138 kg/m. In addition, BC supports a superimposed dead load of 6 kN/m. Assume that both members AB and BC are axially rigid. E = 20,000 MPa.

3.15

A simply supported beam is represented by the lumped mass model shown in Figure P3.15. Obtain the stiffness matrix of the beam for the six degrees-of-freedom shown in the figure. Eliminate the rotational degrees-of-freedom by a static condensation of the stiffness matrix to obtain a two by two matrix, then derive the equations of motion for the free vibrations of the beam with reference to the two translational degrees-of-freedom.

Figure P3.15

Chapter 4

Principles of analytical mechanics

4.1

INTRODUCTION

The development of the science of mechanics has relied on two different sets of fundamental principles. The first of these is based on Newton’s laws of motion. These laws deal with the motion of a particle under the action of forces acting on it, and the quantities of primary interest in the analysis are: the forces, which may include applied forces, forces of interaction between particles, and the forces of constraints; and the momentum. Since both the forces and the momentum are vector quantities, the branch of mechanics based on the principles enunciated by Newton is called vectorial mechanics. When the methods of vectorial mechanics are used for the solution of a problem, the forces of constraints and the forces of interaction must be explicitly accounted for and evaluated even when there is no interest in obtaining their values. The second branch of mechanics, called analytical mechanics, is based on the works of Bernoulli, Euler, d’Alembert, Lagrange, Poisson, Hamilton, Jacobi, and Gauss. As contrasted to force and momentum, the parameters of interest in analytical mechanics are scalar functions, and therefore the fundamental equations used in it do not depend on the choice of coordinates. Also, it is not necessary that the forces of constraints be explicitly accounted for or evaluated. The methods of analytical mechanics are therefore very effective in the solution of complex systems with multiple constraints. This chapter deals with the basic principles of analytical mechanics and their application to the problems of structural dynamics. The concepts of generalized coordinates and generalized forces are introduced at the beginning. The important principle of virtual work is described next. This is followed by a presentation of Hamilton’s principle and Lagrange’s equation of motion. The notations and the methods of variational calculus are used throughout the chapter.

4.2

GENERALIZED COORDINATES

In vectorial mechanics the position of a particle mass in the geometric space is specified by the three rectangular coordinates of Descarte. It is not however, necessary to use the rectangular coordinates; other appropriately selected parameters may be used as effectively to locate the position of the particle. For example, if the particle is constrained to move in a plane, we may use the two polar coordinates, r and θ , shown in

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Dynamics of structures

Figure 4.1 (a) Polar coordinates; (b) spherical coordinates.

Figure 4.1a. The transformation from the polar to the rectangular coordinates x1 and x2 is given by the following equations x1 = r cos θ x2 = r sin θ

(4.1)

In a three-dimensional space, one might use the spherical coordinates r, θ , and φ, shown in Figure 4.1b. The transformation is in this case given by x1 = r sin θ cos φ x2 = r sin θ sin φ

(4.2)

x3 = r cos θ The polar or spherical coordinates are by no means the only types of coordinates that may be used to specify the position of a particle in space. To illustrate the point, consider a particle P moving in a plane (Fig. 4.2). Also consider the line joining a point A having coordinates (2, 4) and a point B having coordinates (4, 2). The particle P can then be located by specifying the angles q1 and q2 that lines AP and BP, respectively, make with line AB. It is easily proved that coordinates sets (q1 , q2 ) and (x, y) are related by the following transformation equations: x=4+2

sin q1 (sin q2 + cos q2 ) sin (q2 − q1 )

y=2+2

sin q1 (sin q2 − cos q2 ) sin(q2 − q1 )

(4.3)

Note that when P lies on the line AB, q1 and q2 are both zero and functions given by Equation 4.3 are no longer single valued. Coordinates q1 and q2 cannot, therefore, be used for describing the motion of P near line AB.

Principles of analytical mechanics 121

Figure 4.2 Generalized coordinates.

The arbitrary coordinates used above to specify the location of a particle are examples of what are called generalized coordinates. Their number is the minimum required to specify the position of a particle or particles in space. In general, a mechanical system may consist of a number of particles. Let this number be N. If there are no constraints, 3N coordinates, rectangular or generalized, will be required to specify the position of the system in space. The 3N rectangular coordinates, x1 , y1 , z1 , x2 , y2 , z2 , . . . , zN , are related to 3N generalized coordinates q1 , q2 , . . . , q3N by equations of the form x1 = f1 (q1 , q2 , q3 , . . . , q3N ) y1 = f2 (q1 , q2 , q3 , . . . , q3N ) z1 = f3 (q1 , q2 , q3 , . . . , q3N ) ... ... .................. zN = f3N (q1 , q2 , q3 , . . . , q3N )

(4.4)

It is necessary that functions given by Equation 4.4 be finite, single valued, continuous and differentiable. In the examples cited so far, the number of generalized coordinates was equal to the number of rectangular coordinates and there was no particular advantage in using the former. However, when constraints exist in a system, the number of generalized coordinates may be significantly less than the number of rectangular coordinates, and the use of the former may considerably reduce the complexity of the problem. As an example of constraints, consider the double pendulum shown in Figure 4.3. The position of the two point masses can be defined by specifying the rectangular coordinates x1 , y1 and x2 , y2 . However, the four coordinates are related by the following two equations of constraint: x21 + y12 = l12 (x2 − x1 )2 + (y2 − y1 )2 = l22

(4.5)

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Dynamics of structures

Figure 4.3 Double pendulum.

Figure 4.4 Cylinder rolling on a plane.

These two constraint equations imply that the four coordinates x1 , y1 and x2 , y2 cannot all be independently specified. In fact, if two of them are given, the other two are determined automatically. It is obvious that the minimum number of coordinates required to specify the position of the system is 2. Although we may choose x1 and x2 as the two generalized coordinates, it is probably easier to use the two angles q1 and q2 shown in Figure 4.3. A particle moving on a plane is constrained so that the displacement normal to the plane is always zero. This condition gives one equation of constraint so that the minimum number of independent coordinates needed to specify the position of the particle is two rather than three. The position of a cylinder rolling on a horizontal plane can be specified by the two coordinates shown in Figure 4.4: the x distance of the center of cylinder from a prescribed vertical reference axis and the angle that a given radius OP makes with the vertical. However, when the cylinder rolls without slipping, these two coordinates are related by one equation of constraint. This equation obtained from the condition that the instantaneous velocity of the point of contact should be zero is rθ˙ = x˙

(4.6)

Principles of analytical mechanics 123

Figure 4.5 Representation of displaced shape through superposition of modes.

Integration of Equation 4.6 yields rθ = x

(4.7)

Coordinates θ and x are called constrained coordinates. The minimum number of coordinates required to specify the position of the cylinder is one; and either x or θ may be chosen as the generalized coordinate. If θ is chosen as the independent coordinate, the transformation equations can be written as x = rθ θ =θ

(4.8)

Based on the foregoing discussion, we can now provide a formal definition for the term “generalized coordinates.’’ The minimum number of independent coordinates required to specify the position of a mechanical system in space are defined as the generalized coordinates. For systems of interest to us, the number of such coordinates is equal to the number of degrees of freedom of the system. Thus a particle moving on a plane has two degrees of freedom while a cylinder rolling on a rough plane has only one degree of freedom. It is obvious that generalized coordinates are not unique; rather, there are an infinite number of different coordinate sets. The choice of a particular set will depend on the ease with which a problem can be formulated in that set. In the examples cited so far, the generalized coordinates were identified with physical quantities that could be measured. This is not always necessary, however, and we may in fact choose generalized coordinates that have no physical meaning. As an example, consider three particle masses attached to a massless flexible cantilever shown in Figure 4.5. If the bar is assumed to be axially rigid, the positions of the particle masses can be determined by specifying the three coordinates x1 , x2 , and x3 . Figure 4.5 also

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Dynamics of structures

shows three possible displaced shapes of the cantilever. The choice of displacement shapes is arbitrary except that they should be independent, so that none of them can be derived by a linear combination of the remaining two. Let these shapes be specified by three vectors φ1 , φ2 and φ3 given by φ1T = [1 2 3] φ2T = [1 0 −1] φ3T = [1 −1 1]

(4.9)

It is possible to describe any displaced shape of the cantilever bar by a superposition of the shape functions φ1 , φ2 and φ3 as follows: 

 x1 x =  x2  = α1 φ1 + α2 φ2 + α3 φ3 x3

(4.10)

or in the equivalent form   1 1 1 α1 x =  2 0 −1   α2  = Cα 3 −1 1 α3 

(4.11)

in which xT = (x1 , x2 , x3 ) and α1 , α2 , and α3 are the coefficients to be determined. For instance, a displacement shape in which x1 = 1, x2 = 2.5 and x3 = 2.0 can equally well be described by specifying α1 = 1, α2 = 12 , and α3 = − 12 . Since vectors φ1 , φ2 and φ3 were chosen to be independent, the transformation matrix C is nonsingular and a unique value of vector α can be determined for every value of x. The coefficients α1 , α2 , and α3 can be interpreted as generalized coordinates. It is not, however, possible to assign any physical meaning to them. Later, in Chapters 10 and 12, it will be shown that the normal coordinate transformation which plays a central role in problems of structural dynamics is indeed a special case of the transformation given by Equation 4.11.

4.3

CONSTRAINTS

The idea of constraints was briefly introduced in the Section 4.2. However, because constraints play an important role in analytical mechanics and the particular method of solution used may depend on the type of constraints in the system, it is of interest to take a closer look at the nature of constraints and the resulting constraint equations. In the following discussion, the symbol u is employed to designate constrained coordinates, while q is used for generalized coordinates. Constraint equations 4.5 and 4.7 are special cases of the general form f (u1 , u2 , . . . , uN , t) = 0

(4.12)

Principles of analytical mechanics 125

Figure 4.6 Thin disk rolling on a plane.

Equation 4.12 prescribes a relationship that involves the constrained coordinates and the time. The equations of constraint may involve differentials of coordinates rather than the coordinates themselves. An example is provided by Equation 4.6, rewritten in the form r dθ − dx = 0

(4.13)

Equation 4.13, which involves differentials of the coordinates, can be converted to the form of Equation 4.12 by simple integration. Such reduction may not, however, always be possible. A constraint of the form of Equation 4.12 or reducible to that form is called a holonomic constraint. A constraint that cannot be reduced to the form of Equation 4.12 is called a nonholonomic constraint. As an example of nonholonomic constraints, consider the motion of a thin disk rolling on a rough plane. As shown in Figure 4.6, the position of the disk can be described by specifying the coordinates of the point of contact A, the angle between a vertical plane and the plane containing the disk, and the angle that the tangent to the circumference of the disk at the point of the contact makes with the y axis. The four coordinates described in the foregoing paragraph are, in fact, independent. This is easily verified by noting that even when any three of them have been prescribed, the fourth one can take any arbitrary value from within its admissible range of values. Thus when x, y, and φ are given, θ can take any value within the range −π/2 to π/2. However, even though the coordinates are independent, a constraint equation must exist. This equation is derived by noting that the instantaneous motion should be along the tangent line OA, so that x˙ = y˙ tan φ

(4.14a)

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Dynamics of structures

Figure 4.7 Pendulum of variable length.

or dx = dy tan φ

(4.14b)

Since tan φ is also a variable, it is not possible to integrate Equation 4.14 to reduce it to the form of Equation 4.12; the constraint described by Equation 4.14 is therefore nonholonomic. This constraint reduces the number of degrees of freedom of the system to three, which is one less than the number of independent coordinates. Thus, for systems having nonholonomic constraints, the number of degrees of freedom is less than, and not equal to, the number of generalized coordinates. In a majority of engineering problems, the constraints are holonomic. Only systems with holonomic constraints will therefore be treated in this book. Holonomic constraints can be further subdivided into two categories: rhenomic and scleronomic. When a constraint equation contains time t explicitly, the constraint is said to be rhenomic, when time t is not present, the constraint is scleronomic. The holonomic constraints described by Equations 4.5 and 4.7 are both scleronomic. As an example of a rhenomic constraint, consider the simple pendulum shown in Figure 4.7 in which the string from which the mass is suspended is being pulled upward with a constant velocity v. Let the length of string at time t = 0 be l. Also let x and y be the rectangular coordinates describing the position of point mass m, and let θ be the angle that the string makes with the vertical. The rectangular coordinates are not independent but satisfy a constraint equation of the form x2 + y2 = (l − vt)2

(4.15)

Equation 4.15 contains time t explicitly and describes a rhenomic holonomic constraint. Note that angles θ can be used as the generalized coordinates describing the

Principles of analytical mechanics 127

system. The transformation between the rectangular and the generalized coordinate θ is given by x = (l − vt) sin θ y = (l − vt) cos θ

(4.16)

4.4 VIRTUAL WORK If a particle is at rest under the action of a set of forces, Newton’s second law requires that the vector sum of all the forces acting on the particle be zero. Mathematically, this can be stated as F=0

(4.17)

where F represents the vector sum of forces. In general, the forces acting on the particle can be categorized as applied forces and the forces of constraints. As an example, if the particle is constrained to move on a plane, the reaction normal to the plane is classified as a force of constraint. If the vector sum of applied forces is denoted by Fa and that of the forces of constraints by Fc , Equation 4.17 can be rewritten in the form Fa + F c = 0

(4.18)

Now suppose that the particle in equilibrium is given a virtual displacement δr. The term virtual displacement signifies that the displacement is imaginary and not due to any real cause. We use the symbol δr to denote such an imaginary displacement. The work done by the real force in riding through the imaginary displacement is termed virtual work and is given by δW = F · δr

(4.19)

where a dot represents the scalar product of two vectors. Again, we use the symbol δW rather than W or dW to signify that the work done is not performed by real forces undergoing real displacements. Since F is zero, δW should also be zero, giving δW = 0

(4.20)

Equation 4.20 can be treated as an expression of the principle of virtual work. In this form it is simply a restatement of the equations of static equilibrium. But now let the arbitrary virtual displacement δr be chosen such that the forces of constraint do not work in moving through δr. Such a virtual displacement is said to be compatible with the constraints of the system. For example, in the case of a particle moving on a plane, a virtual displacement parallel to the plane is compatible with the constraint, and the normal reaction will do no work in moving through such a displacement. The virtual work equation can then be written as δW = Fa · δr + Fc · δr = 0

(4.21)

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Dynamics of structures

and since Fc · δr is zero, this gives δW = Fa · δr = 0

(4.22)

Equation 4.22 does not contain the forces of constraint. Also, Fa is no longer zero; only the scalar product of Fa and δr is zero. In this form, Equation 4.22 is quite different from Equation 4.17. To illustrate the points mentioned above, consider again the case of a particle moving in the xy plane. The particle is in equilibrium under an applied force and the reaction exerted by the plane. Let the components of applied force be Fx , Fy , and Fz , and let the reaction normal to the plane be denoted by N. For an arbitrary displacement with component δx, δy and δz, the virtual work equation becomes Fx δx + Fy δy + (Fz + N)δz = 0

(4.23)

Now because δx, δy, and δz are independent, Equation 4.23 is entirely equivalent to the three equations Fx = 0, Fy = 0, and (Fz + N) = 0, which are readily seen to be the equations of equilibrium. If the virtual displacement is consistent with the constraint, component δz should be zero and Equation 4.23 reduces to Fx δx + Fy δy = 0

(4.24)

Equation 4.22 is easily extended to a system consisting of a number of particles. Consider, for example, the case of two particles moving on the xy plane. The virtual work equation is F1 · δr1 + F2 · δr2 = 0

(4.25)

When the virtual displacements are consistent with the constraints, δr1 and δr2 will have no components normal to the plane of motion. Equation 4.25 can therefore be written as Fx1 δx1 + Fy1 δy1 + Fx2 δx2 + Fy2 δy2 = 0

(4.26)

The virtual displacement components in Equation 4.26 are all independent; Equation 4.26 is therefore equivalent to four separate equations which can easily be identified as the equations of equilibrium, two per particle. When additional constraints are introduced, the number of independent virtual displacement components will be reduced correspondingly. For example, if the two particles moving in the xy plane are tied together by a rigid bar, the number of independent virtual displacements is reduced to three. As shown in Figure 4.8, the virtual displacements can be expressed in terms of δq1 , and δq2 , the x and y displacements of the center of the rod, respectively, and δq3 , the rotation of the rod about its center. Since a rigid body is simply a collection of an infinite number of particles constrained to move together, the principle of virtual work is easily extended to such a body. A rigid body moving in a plane has three independent displacement components, two translations and one rotation. A rigid body moving in space has three independent translation components and three independent rotational components.

Principles of analytical mechanics 129

Figure 4.8 Virtual displacements of two interconnected particles constrained to lie on a plane.

Figure 4.9 Two particles connected by a spring lying on the x axis and in equilibrium under forces F 1 and F 2 .

A system consisting of an assemblage of particles and rigid bodies is referred to as a mechanical system. The principle of virtual work for such a system can be stated as follows. If a mechanical system in equilibrium under a set of forces is given an arbitrary virtual displacement that is compatible with the constraints on the system, the sum of work done by the applied forces in undergoing the virtual displacement is zero. The principle of virtual work can also be extended to deformable bodies, but now we must take account of the work done by the internal forces in riding through the virtual deformation. To illustrate the point, consider the simple example of two particles attached to each other by a spring of stiffness k and constrained to move along the x axis as shown in Figure 4.9a. Let the system be in equilibrium under applied forces F1 and F2 shown in the figure, and let the spring be stretched by x under the action of these forces. If the system is given virtual displacements δx1 and δx2 , the virtual work done by the applied forces F1 and F2 will be δWe = F1 δx1 + F2 δx2

(4.27)

Now, although δx1 and δx2 are independent, the virtual work obtained from Equation 4.27 is no longer zero. However, if we add to δWe the work done by the spring force, the total work so obtained will be zero. The forces exerted by the spring on the

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Dynamics of structures

two particles are indicated in Figure 4.9b. The work done by the spring forces, called the internal work, is denoted by δU and is given by δU = −kxδx1 + kxδx2

(4.28)

where x is the stretch of the spring. The modified virtual work equations becomes δWe + δU = 0

(4.29)

(F1 − kx)δx1 + (F2 + kx)δx2 = 0

(4.30)

or

Because δx1 and δx2 are independent, Equation 4.30 is equivalent to the two equations F1 = kx and F2 = −kx. These relationships imply that for equilibrium F1 = −F2 and that at the position of rest the spring will be stretched by x = F1 /k. As an alternative to defining the internal work as in Equation 4.28, we may identify the internal forces acting on the spring rather than on the two particles and calculate the work done on the internal elements. We shall denote it by δWie . For our example, δWie = kx(δx1 − δx2 )

(4.31)

It is obvious that δWie = −δU

(4.32)

On substituting Equation 4.32 into Equation 4.29, the virtual work equation can be written in the alternative form δWe = δWie

(4.33)

In the example that we just considered, the deformable body was a simple spring and the calculation of the internal work was quite straightforward. In general, the deformable bodies will be of more complex shape. However, internal virtual work expressions for elastic bodies of various shapes, including bars, beams, and plates, are derived in standard textbooks on applied mechanics. The principle of virtual work for deformable bodies can now be stated as follows. If a deformable system in equilibrium under a set of forces is given virtual displacements that are consistent with the constraints on the system, the sum of the internal virtual work and the virtual work done by the applied forces is zero. The principle of virtual work occupies the same position in analytical mechanics as Newton’s second law does in vectorial mechanics. They are both fundamental postulates on which all further developments in the respective branches of mechanics are based. One advantage of using the methods of analytical mechanics is at once obvious. Vectorial mechanics requires that the forces acting on each individual particle of the system, including those due to constraints, be identified and evaluated. In analytical mechanics, the forces of constraints need not be accounted for explicitly. This advantage will be evident from the example that follows.

Principles of analytical mechanics 131

Example 4.1 A uniform bar of mass m and length l is constrained to move in a vertical plane by guides as shown in Figure E4.1a. A mass M is attached to one end of the bar and the other end is restrained by a spring of stiffness k. The spring is unstretched when the bar is horizontal. Find the position of equilibrium of the bar.

Figure E4.1 Rigid bar moving in guides; (b) free-body diagram of bar in (a).

Solution The free-body diagram shown in Figure E4.1b indicates all the forces acting on the bar. These forces include the two reactions R1 and R2 . When the virtual work method is used to solve the problem, it is not necessary to evaluate these reactions. The system has a single degree of freedom and its position is completely described by one generalized coordinate. We may choose any one of the coordinates θ, x1 , y1 , and y2 shown in Figure E4.1a. Let angle θ be chosen as the generalized coordinate. A small change δθ in this angle will serve as the virtual displacement. The corresponding variations in x1 , y1 , and y2 are evaluated as follows: l (1 − cos θ ) 2 l y1 = ( sin θ ) 2 y2 = l sin θ

x1 =

(a)

132

Dynamics of structures l sin θ δθ 2 l δy1 = cos θ δθ 2 δy2 = l cos θ δθ

δx1 =

(b)

The equation of virtual work becomes l l l δW = −k (1 − cos θ ) sin θ δθ + mg cos θ δθ + Mgl cos θ δθ = 0 2 2 2

(c)

Equation c leads to the transcendental equation mg kl (1 − cos θ) tan θ = + Mg 4 2

(d)

the solution of which gives the value of angle θ.

4.5

GENERALIZED FORCES

Just as the position of a system can be described in terms of generalized coordinates, the forces acting on the system can be represented by generalized forces. These forces are entirely equivalent to the real forces acting on the system in that the work done by the real forces on a set of virtual displacements is equal to the work done by the generalized forces on the corresponding virtual displacements along the generalized coordinates. This last statement permits us to express the generalized forces in terms of the real forces acting on the system. Suppose that the system subjected to real forces Fi (i = 1, 2, . . . , M) is given virtual displacements δri (i = 1, 2, 3, . . . , M) The virtual work is then δW =

M

Fi · δri

(4.34)

i=1

The overbar on W means that it is simply infinitesimal work, not the variation of a scalar function W. Now the radius vectors r are related to the N generalized coordinates q1 , q2 , . . . , qN by equations of the form r1 = f1 (q1 , q2 , . . . , qN ) r2 = f2 (q1 , q2 , . . . , qN ) ... ... ............... rM = fM (q1 , q2 , . . . , qN )

(4.35)

Principles of analytical mechanics 133

It follows that δri =

N

∂ri δqj ∂qj

(i = 1, 2, . . . , M)

(4.36)

j=1

The equations of virtual work can therefore be written as

δW =

M

i=1

=



 N

∂r i Fi · δqj  ∂qj j=1

!

N M



j=1

i=1

∂ri Fi · ∂qj

" δqj

(4.37)

By the definition of generalized forces the virtual work should also be given by δW =

N

Qj δqj

(4.38)

j=1

where Qj are the generalized forces. Equations 4.37 and 4.38 should be entirely equivalent. Also, since δqj ’s are all independent, they can be assigned arbitrary values. Comparing Equations 4.37 and 4.38, and setting all δqj ’s except the mth to zero, Qm =

M

i=1

Fi ·

∂ri ∂qm

(4.39)

giving an expression for the mth generalized force. If the system is in equilibrium, δW should be zero, and since δqj ’s are independent, it follows from Equation 4.38 that each Qj must be zero. Unlike real forces, it is not always possible to assign a physical meaning to the generalized forces. Also, the latter need not have the units of a force. However, the products of a generalized force and the corresponding generalized displacement will always have the unit of work.

Example 4.2 A counterbalanced door of weight P rotates freely about a frictionless hinge as shown in Figure E4.2. The balancing weight Q is attached to the tip of the door by a string as shown. Determine the position of the door when it is in equilibrium.

Solution The system has a single degree of freedom and needs one coordinate to describe its position. The angle that the door makes with the vertical is selected as the generalized coordinate. The

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Dynamics of structures

Figure E4.2 Counterbalanced door. constrained coordinates y1 and y2 are related to the generalized coordinates by the following equations y1 = a cos θ j 
θ j y2 = C + 4a sin 2

(a) (b)

where C is a constant and j is a unit vector along the y axis. The virtual work done by the real forces −Qj and −Pj is given by δW = −Qj · δy2 − Pj · δy1 ∂y ∂y1 = −Qj · 2 δθ − Pj · δθ ∂θ ∂θ   
θ + Pa sin θ δθ = −Q 2a cos 2

(c)

In terms of generalized forces, the virtual work is δW = Qθ δθ

(d)

Comparison of Equations c and d gives the generalized force Q: Qθ = Pa sin θ − 2aQ cos

θ 2

(e)

The condition that Qθ should be zero for equilibrium gives the following transcendental equation in θ : θ P sin θ = 2Q cos 2

Principles of analytical mechanics 135 or

θ Q = 2 P Note that the generalized force Qθ has the units of a moment. sin

Example 4.3 A rigid uniform bar of mass m supported at its ends by springs of stiffnesses k1 and k2 is subjected to a force P as shown in Figure E4.3a. If the unstretched length of each spring is d, find the position of equilibrium.

Figure E4.3 Rigid bard supported on springs.

Solution Figure E4.3b shows a free-body diagram of the rod. A displacement coordinate is defined along each of the four forces acting on the bar. These coordinates designated y1 , y2 , y3 , and y4 are obviously not independent. The system has two degrees of freedom and therefore needs only two generalized coordinates to specify its position in space. The vertical translation q1 of the center of gravity and the rotation q2 about it are selected as the two generalized coordinates. The constrained coordinates can be expressed in terms of q1 and q2 as follows
 L y1 = q1 − q2 j 2   
L − a q2 j y2 = q1 − 2 y3 = q1 j
 L y4 = q1 + q2 j 2

(a)

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Dynamics of structures

The virtual work equation is given by


 ∂yi ∂yi δq1 + δq2 ∂q1 ∂q2 i=1 


   L L L − a δq2 δq1 − δq2 + P δq1 − = −k1 q1 − q2 2 2 2

 L L δq1 + δq2 −mg δq1 − k2 q1 + q2 2 2

δW =

4

Fi ·

On collecting terms, we obtain 
L L δW = −k1 q1 + k1 q2 + P − mg − k2 q1 − k2 q2 δq1 2 2
 2 PL L L L L2 + k1 q1 − k1 q2 − + Pa − k2 q1 − k2 q2 δq2 2 4 2 2 4

(b)

The virtual work equation in terms of generalized forces is δW = Q1 δq1 + Q2 δq2

(c)

On comparing Equations b and c, we have L Q1 = −(k1 + k2 )q1 + (k1 − k2 ) q2 + (P − mg) 2 
2 L L l (k1 + k2 )q2 + P a − Q2 = (k1 − k2 )q1 − 4 2 2

(d)

For equilibrium, each of Q1 and Q2 should be zero. This condition leads to the following equations in q1 and q2 :    k1 + k2 − L2 (k1 − k2 ) P − mg q1 = (e) 2 q2 P(a − L2 ) − L2 (k1 − k2 ) L4 (k1 + k2 ) The solution of Equation e will give q1 and q2 . Note that Q1 has the units of force, while Q2 has the units of a moment.

Example 4.4 The vertical cantilevers of Figure E4.4 is subjected to the forces as shown. If the coordinates described by Equation 4.11 are used as the generalized coordinates, find the corresponding generalized forces.

Solution The virtual work equation is given by δW = P1 δx1 + P2 δx2 + P3 δx3    δα1 1 1 1 0 −1   δα2  = [ P1 P2 P3 ]  2 3 −1 1 δα3

 δα1 = [(P1 + 2P2 + 3P3 ) (P1 − P3 ) (P1 − P2 + P3 )]  δα2  δα3 

(a)

Principles of analytical mechanics 137

Figure E4.4 Vertical cantilever.

from which the three generalized forces are readily identified as Q1 = P1 + 2P2 + 3P3 Q2 = P1 − P3

(b)

Q3 = P1 − P2 + P3 Note that in this case it is not possible to give a physical meaning to the generalized forces, although they all have the units of force.

4.6

CONSERVATIVE FORCES AND POTENTIAL ENERGY

Consider a particle moving in space under the action of a force F as shown in Figure 4.10. The work done by the force F in moving from point A to B is given by  W=

B

F · dr

(4.40)

A

where r denotes the radius vector. In general, the work done will depend on the path the particle follows from A to B. However, as a special case, consider a force that can be derived from a scalar function φ as follows

F=

∂φ ∂φ ∂φ i+ j+ k ∂x ∂y ∂z

= ∇φ

(4.41)

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Dynamics of structures

in which the operator ∇, called the gradient vector, is given by ∇=

∂ ∂ ∂ i+ j+ k ∂x ∂y ∂z

(4.42)

On substituting Equation 4.41 into Equation 4.40 and noting that dr = dxi + dyj + dzk, 

B

W= 




A

∂φ ∂φ ∂φ dx + dy + dz ∂x ∂y ∂z



B

=

dφ A

= φ B − φA

(4.43)

Equation 4.43 shows that the work done in moving the particle from point A to B depends only on the values of the function φ at those points and is independent of the path along which the particle moves. When the work done by a force during the motion of its point of application is independent of the path followed but depends only on the position of the terminal points of the path, that force is said to be conservative. A force that can be derived from a scalar function according to Equation 4.41 is thus a conservative force. The work done by a conservative force in moving from a point A to datum O is designated as the potential energy of the force at A. Thus  VA =

O

F · dr

(4.44)

A

where V denotes the potential energy. Now let the particle move from A to B along a path that includes the datum O (Fig. 4.10). Then, since the work done is independent

Figure 4.10 Work done by a force in moving along a given path.

Principles of analytical mechanics 139

of the path, Equation 4.40 can be written as  W=  =

O

 F · dr +

A O

 F · dr −

A

B

F · dr

O O

F · dr

B

= −(VB − VA )

(4.45)

Equation 4.45 implies that the work done is the negative of the change of potential energy. In differential form this can be stated as dW = dφ = −dV

(4.46)

implying that dW is a complete differential. From Equations 4.46 and 4.41 it follows that F = −∇V

(4.47)

As an example of a conservative force, consider the gravitational force near the surface of the earth. As shown in Figure 4.11, the force acting on a particle of mass m is given by F = −mgk

(4.48)

Now define a scalar function φ = −mgz. It is seen that ∇φ = −mgk = F

(4.49)

The gravitational force can thus be expressed as the gradient of a scalar function and is therefore conservative. If the surface of the earth is taken as the datum, the potential energy of the mass at a height z above the surface is given by VA = mgz

(4.50)

Figure 4.11 Gravitational force of the earth.

140

Dynamics of structures

Figure 4.12 Work done against the force of a spring.

The force exerted by a linear spring of stiffness k is another example of a conservative force (Fig. 4.12). The work done by the spring force when the end of the spring is moved a distance x from the unstretched position, is given by 

x

W= 0

1 −kx dx = − kx2 2

(4.51)

The negative of this work is the potential energy stored in the spring. Thus V=

1 2 kx 2

(4.52)

and F=−

dV = −kx dx

(4.53)

The energy stored in an elastic body when it is deformed is called strain energy. Expressions for strain energies stored due to flexure, axial, shear, and torsional deformations can be found in any standard textbook on applied mechanics. There are many forces that are nonconservative in nature and cannot be derived from a scalar function. The force of friction is an example of a nonconservative force, since the work done by it is not independent of the path through which its point of application moves. Example 4.5 (a) (b)

Find the conditions under which a force F = Ai + Bj + Ck is conservative. A particle moving in space is subjected to a force of constant magnitude, always directed toward the origin. Using the result obtained in part (a), show that the force is conservative.

Solution (a)

The infinitesimal work is given by dW = F · dr = A dx + B dy + C dz

(a)

If the force F is conservative, then from Equation 4.46, dW = dφ. In other words, dW must be a complete differential. From elementary calculus, we know that for dW to be a complete

Principles of analytical mechanics 141 differential, the following conditions must be satisfied ∂A ∂B − =0 ∂y ∂x ∂A ∂C − =0 ∂z ∂x ∂B ∂C − =0 ∂z ∂y (b)

(b)

A force of magnitude G directed toward the origin is given by −G

y z  i+ j+ k r r r

x

(c)

where r = (x2 + y2 + z2 )1/2 Setting A = −Gx/r, B = −Gy/r, and C = −Gz/r gives us −xy yx ∂A ∂B − = 2 + 2 ∂y ∂x (x + y2 + z2 )2/3 (x + y2 + z2 )2/3 =0

(d)

In a similar manner, it can be proved that the other two conditions specified in Equation b are also satisfied. The force F is therefore conservative. A force that is always directed toward a fixed point in space, as shown in Figure E4.5, is called a central force. It can be shown that if the magnitude of the central force is proportional to rn , the force is conservative. Gravitational forces between two bodies and the force between two electric charges, both proportional to 1/r2 , are examples of conservative central forces.

Figure E4.5 Central force.

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Dynamics of structures

4.7 WORK FUNCTION Equation 4.38 is an expression of the work done when a system is subjected to infinitesimal virtual displacements δqj ’s. Now, among the set of admissible virtual displacements are also the differentials dqj ’s. On replacing displacements δqj ’s by dqj ’s the expression for the infinitesimal work becomes dW = Q1 dq1 + Q2 dq2 + · · · + QN dqN

(4.54)

Now, if a scalar function U can be found such that ∂U = Qi ∂qi

i = 1, 2, . . . , N

(4.55)

then the right-hand side of Equation 4.54 is equal to dU and dW becomes a complete differential dW = dU

(4.56)

Following the reasoning advanced in Section 4.6, it can be shown that because the infinitesimal work dW is a complete differential, the generalized forces, Qi , are conservative. The scalar function U is called the work function and its negative is equal to the potential energy V, so that U(q1 , q2 , q3 , . . . , qN ) = −V(q1 , q2 , q3 , . . . , qN )

(4.57)

and −

∂V = Qi ∂qi

(4.58)

Further, if the system is in equilibrium, then as proved in Section 4.5, all generalized forces must be equal to zero and the following relations hold ∂V =0 ∂qi

i = 1, 2, . . . , N

(4.59)

Conditions expressed by Equation 4.59 are precisely the conditions for a stationary value of the function V. The stationary value may represent a minimum, a maximum, or a saddle point. It can be shown that a minimum value represents a state of stable equilibrium, a maximum value represents a state of unstable equilibrium, and a saddle point corresponds to a position of neutral equilibrium. Example 4.6 For the counterbalanced door of Example 4.2, determine the position of equilibrium by finding a stationary value of the potential energy. Is the equilibrium stable?

Principles of analytical mechanics 143

Solution Ignoring a constant additive, the potential energy is given by V = 4aQ sin

θ + Pa cos θ 2

(a)

This potential energy function can also be obtained from the expression for infinitesimal virtual work derived in Example 4.2 using the reasoning outlined in Section 4.7. Thus   
θ + Pa sin θ dθ dW = −Q 2a cos 2

(b)

Hence 
θ Qθ = −Q 2a cos + Pa sin θ 2

(c)

It is readily seen that the work function U which satisfies the condition ∂U/∂θ = Qθ is given by


θ U = −Q 4a sin 2

 − Pa cos θ

(d)

The forces acting on the door are thus conservative and the potential energy function which is negative of the work function is given by Equation a. For a stationary value of the potential energy, ∂V =0 ∂θ which gives 2aQ cos

θ − Pa sin θ = 0 2

(e)

or sin

θ Q = 2 P

(f)

Equation f shows that equilibrium is possible if Q, the counterbalancing weight is less than P, the weight of the door. To determine whether equilibrium is stable, find ∂ 2 V/∂θ 2 at the position of equilibrium. θ ∂ 2V = −aQ sin − Pa cos θ ∂θ 2 2

(g)

On substituting the value of sin(θ/2) from Equation f, Equation g gives
 ∂ 2V Q2 Q − Pa 1 − 2 = −aQ P P2 ∂θ 2 a = (Q2 − P2 ) P

(h)

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Dynamics of structures

Figure E4.6 Variation of potential energy. For Q less than P, Equation h will always give a negative value for ∂ 2 V/∂θ 2 . This means that the position of equilibrium corresponds to the maximum value of the potential energy and the equilibrium is unstable. This can be verified from the graph of potential energy versus angle θ plotted in Figure E4.6 for Q/P = 1/2.

Example 4.7 A rigid bar of mass m and length l is connected at one end to another rigid bar, also of mass m and length l. The free end of the first bar is attached to the ceiling by a hinge. If a horizontal force P acts on the free end of the second bar, find the position of equilibrium of the system.

Solution The system has two degrees of freedom and can be described by the generalized coordinates q1 and q2 as shown in Figure E4.7. If the potential energy is assumed to be zero when the parts are in a vertical position, the potential energy for any other position can be expressed in terms of q1 and q2 as follows V=

  l l mg(1 − cos q1 ) + mg l(1 − cos q1 ) + (1 − cos q2 ) − Pl(sin q1 + sin q2 ) 2 2

(a)

The conditions for the stationary value of V will give the position of equilibrium as follows ∂V =0 ∂q1 or l mg sin q1 + lmg sin q1 − Pl cos q1 = 0 2 2 P tan q1 = 3 mg

(b)

Principles of analytical mechanics 145

Figure E4.7 Multiple links under equilibrium. and δV =0 δq2 l mg sin q2 − Pl cos q2 = 0 2 2P tan q2 = mg

4.8

(c)

LAGRANGIAN MULTIPLIERS

In the solution of a problem it may at times be simpler to work with constrained coordinates. The differential work Equation 4.54 when written in the M constrained coordinates becomes dW = F1 du1 + F2 du2 + · · · + FM duM

(4.60)

If forces F1 , F2 , . . . , FM are conservative, Equation 4.60 reduces to dW = −

∂V ∂V ∂V du1 − du2 − · · · − dum ∂u1 ∂u2 ∂um

(4.61)

in which the overbar on dW has been removed because dW is now a complete differential. For equilibrium, dW must be zero. However, since the coordinates u1 , u2 , . . . , uM are not independent, Equation 4.61 does not imply that the coefficients of the differential du1 , du2 , . . . , duM are zero. If the number of generalized coordinates is N, where N < M, there must exist (M − N) = L constraint equations of the form φ1 (u1 , u2 , . . . , uM ) = 0 φ2 (u1 , u2 , . . . , uM ) = 0 ······················ φL (u1 , u2 , . . . , uM ) = 0

(4.62)

146

Dynamics of structures

Equation 4.62 can be written in the alternative form ∂φ1 ∂φ1 du1 + du2 + · · · + ∂u1 ∂u2 ∂φ2 ∂φ2 du1 + du2 + · · · + ∂u1 ∂u2

∂φ1 duM = 0 ∂uM ∂φ2 duM = 0 ∂uM

···································· ∂φL ∂φL ∂φL du2 + · · · + duM = 0 du1 + ∂u1 ∂u2 ∂uM

(4.63)

If each of Equations 4.63 is multiplied by an arbitrary function, the multiplier for the ith equation being denoted by λi , and the resulting products are added to Equation 4.61, then at equilibrium the following equation is obtained.


 ∂φ1 ∂φL ∂V du1 + λ1 + · · · + λL − ∂u1 ∂u1 ∂u1 
∂φ1 ∂φL ∂V du2 + − + λ1 + · · · + λL ∂u2 ∂u2 ∂u2 
∂φ1 ∂φL ∂V duM = 0 +··· + − + λ1 + · · · + λL ∂uM ∂uM ∂uM

(4.64)

Now since λ1 , λ2 , . . . , λL are arbitrary, they can be so chosen that the expressions in the first L set of parentheses of Equation 4.64 are zero. This will leave M − L = N expressions in parentheses in that equation, but since the system has N degrees of freedom, the coefficients of these expressions dui can be assigned arbitrarily and it follows that the expressions in these remaining sets of parentheses should also each be zero. The foregoing reasoning leads to M equations, which together with the L constraint conditions will permit a solution for the M displacement coordinates u1 , u2 , . . . , uM and the L multipliers, λ1 , λ2 , . . . , λL . The arbitrary multipliers used in the method above are called Lagrangian multipliers. Example 4.8 A massless rigid bar is suspended from the ceiling by three wires as shown in Figure E4.8. The elastic properties of the wires are 1/2

T1 = b1 u1 T2 = b2 u2 T3 = b3 u3

where T represents tension in a wire, u the displacement of the end of the wire, and b a material property constant. If the bar supports a central vertical load F, find the tension in each wire.

Principles of analytical mechanics 147

Figure E4.8 Rigid bar suspended by three wires.

Solution The system has two degrees of freedom, but the three coordinates u1 , u2 , and u3 will be used in the solution of the problem. These coordinates satisfy a constraint equation of the form φ = u1 − 2u2 + u3 = 0

(a)

The potential energy of the system comprises the strain energy stored in the wires and the potential energy of the load. It can easily be shown to be given by V=

2 1 1 3/2 b1 u1 + b2 u22 + b3 u23 − Fu2 3 2 2

(b)

The virtual work equation can be written as dW = −

∂V ∂V ∂ du1 − du2 − du3 = 0 V∂u1 ∂u2 ∂u3

or 1/2

b1 u1 + (b2 u2 − F)du2 + b3 u3 du3 = 0

(c)

The virtual work equation can also be obtained directly as dW = −T1 du1 − T2 du2 − T3 du3 + F3 du2 = 0 which will lead to Equation c. The differentials du1 , du2 , and du3 in Equation c are not independent since u1 , u2 , and u3 are related by constraint equation a. That equation can be expressed as dφ =

∂φ ∂φ ∂φ du1 + du2 + =0 ∂u1 ∂u2 ∂u3

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Dynamics of structures

or du1 − 2du2 + du3 = 0

(d)

On multiplying Equation d by an arbitrary multiplier λ and adding to Equation c, we obtain 1/2

(b1 u1 + λ)du1 + (b2 u2 − F − 2λ)du2 + (b3 u3 + λ)du3 = 0

(e)

If λ is chosen so that the expression in the first set of parentheses in Equation e is zero, that equation is left with only two differential coordinates: du2 and du3 . Since the system has two degrees of freedom, these differentials can be assigned arbitrary values. It follows that the expressions in the remaining two sets of parentheses in Equation e should also be zero. We thus obtain the following equations 1/2

b 1 u1 + λ = 0 b2 u2 − F − 2λ = 0

(f)

b3 u3 + λ = 0 Solution of Equations f along with Equation a gives   %
2 F  b2 + 4b3 b21  b2 + 4b3 − +8 2 λ= 2 b 2 b3 b2 b3 b 1 b2 T1 = −λ

(g)

T2 = F + 2λ T3 = −λ

4.9 VIRTUAL WORK EQUATION FOR DYNAMICAL SYSTEMS The principal of d’Alembert, described in Chapter 2, converts a problem of dynamics into an equivalent problem of static equilibrium. This means that the methods of analytical mechanics used for static problems can be applied to dynamical problems as well. For a system composed of M particles, the principle of d’Alembert can be stated as Fi − mi r¨i = 0

i = 1, . . . , M

(4.65)

where Fi is the vector sum of the impressed forces and the forces of constraint acting on particle i, and mi r¨i is the force of inertia. At a particular instant of time, Equation 4.65 represents an equation of equilibrium. Therefore, if the system is given virtual displacements δri which are compatible with the constraints at that instant of time, the virtual work equation can be written as M

i=1

(Fai + Fci − mi r¨i ) · δri = 0

(4.66)

Principles of analytical mechanics 149

Figure 4.13 Pendulum of variable length: (a) virtual displacement; (b) real displacement.

where Fai is the vector sum of the impressed forces and Fci is the vector sum of the forces of constraints. Since virtual displacements do no work on the forces of constraints, Equation 4.66 becomes M

(Fai − mi r¨i ) · δri = 0

(4.67)

i=1

It should be noted that virtual displacements do not occupy any time. In other words, the mathematical experimentation in which the system is given the virtual displacement takes place while the system is frozen in its state described by Equation 4.65. The real displacements, on the other hand, take place over real time and may no longer be admissible as virtual displacements. If that is so, δri ’s in Equation 4.67 cannot be replaced by dri ’s. As an example of a situation in which the real displacements are not admissible as virtual displacements, consider the simple pendulum shown in Figure 4.13 in which the string is being pulled up at a constant velocity so that the distance from the point of suspension to the point mass reduces with time. The position of the pendulum at a certain instance of time is shown in Figure 4.13a. The virtual displacement δr given to the system at that time is along the tangent to an arc with center O and radius OA. Since this displacement is perpendicular to the force of constraint exerted by the string, the string tension does no work on it. The real displacement that takes place in a short time is shown as AB in Figure 4.13b. This displacement is not perpendicular to OA and is therefore not admissible as a virtual displacement. The force of constraint will, in fact, do work on the real displacement. The constraint condition for the pendulum of Figure 4.13 as given by Equation 4.15 is rhenomic because it contains time t explicitly. Whenever constraints are rhenomic, infinitesimal real displacements are not admissible as virtual displacements.

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Dynamics of structures

Consider now a situation in which the constraints are scleronomic, so that dri ’s become admissible virtual displacements. Equation 4.67 can then be written as M

(Fai − mi r¨i ) · dri = 0

(4.68)

i=1

The second term in the expression above can be reduced as follows M

 M
d 1 mi r˙i · r˙i dt dt 2 i=1 " !M d 1 2 = mi r˙i dt dt 2

mi r¨i · dri =

i=1

i=1

= dT

(4.69)

where T is the kinetic energy of the system. If the impressed forces can be derived from a potential energy function V of the type V = V(r1 , r2 , r3 , . . . , rM ), M

Fai · dri =

i=1

M

−

i=1

∂V · dri ∂ri

= −dV

(4.70)

Note that if V contains t explicitly, Equation 4.70 no longer holds good because then dV =

∂V ∂ri

· dri +

∂V dt ∂t

(4.71)

As an example, the pendulum of Figure 4.13 has a potential energy function given by V = l − (l − vt) cos θ

(4.72)

which contains time explicitly. Equation 4.70 is not applicable to this pendulum. Substitution of Equations 4.69 and 4.70 into Equation 4.68 gives −(dV + dT) = 0

(4.73a)

V + T = constant

(4.73b)

or

Equation 4.73 states that in a system which is scleronomic in both the work function and the constraints, the sum of the potential and the kinetic energies is a constant. This principle of conservation of energy, although not sufficient for obtaining a solution to multi-degree-of-freedom systems, does give a complete solution for a single-degree-offreedom system.

Principles of analytical mechanics 151

Example 4.9 A uniform rigid bar of mass M and length l is pinned at one end to a support and at the other end to a uniform cylinder of radius r and mass m as shown in Figure E4.9. The cylinder rolls without slipping inside a cylindrical surface of radius l + r. Find the equation of motion.

Figure E4.9 Cylinder rolling inside another cylinder.

Solution If φ is the absolute rotation of the cylinder (see Figure E4.9), then since the cylinder rolls without slip, r(θ + φ) = (l + r)θ rφ = lθ

(a)

The potential energy of the system measured from the vertical position is V=

l (1 − cos θ)Mg + lmg(1 − cos θ) 2

(b)

The kinetic energy is given by 1 Ml 2 2 1 mr2 2 1 2 2 ˙ + ml (θ˙ ) (θ˙ ) + (φ) 2 3 2 2 2 1 2 2 3 2 2 = Ml θ˙ + ml θ˙ 6 4

T=

(c)

Since the system is scleronomic, the energy is conserved so that d (V + T) = 0 dt or


 
2 Mgl d θ˙ Ml 3 + mgl sin θ θ˙ + + ml 2 θ˙ =0 2 3 2 dt

(d)

152

Dynamics of structures

Since this would be true for all values of t, θ˙ can be canceled out, giving the following equation of motion


 
Ml 2 3 Mgl + ml 2 θ¨ + + mgl sin θ = 0 3 2 2

(e)

When the system executes small oscillations, sin θ in Equation e can be replaced by θ. A possible solution to the equation of motion is then θ = A sin ωt

(f)

where ω, called the frequency of oscillation, obtained by substituting for θ and θ¨ from Equation f into Equation e, is % ω=

g(m + M/2) l(M/3 + 3m/2)

(g)

Example 4.10 The system shown in Figure E4.10 is mounted in a vertical plane. It is set vibrating about its position of equilibrium. Find the equations of motion. The spring is unstretched when the bar is in a horizontal position.

Figure E4.10 Vibrations of rigid bar moving in guides.

Solution The potential energy of the system is given by V=

1 2 kx − mgy1 − Mgy2 2 1

(a)

Principles of analytical mechanics 153 and the kinetic energy is T=

1 2 1 ml 2 2 1 1 m˙y + θ˙ + My˙ 22 + Mx˙ 22 2 1 2 12 2 2

(b)

Referring to Figure E4.10, we see that l (1 − cos θ ) 2 l x2 = (1 + cos θ ) 2 l y1 = sin θ 2 y2 = l sin θ x1 =

(c)

From Equation c, l x˙ 1 = + sin θ θ˙ 2 l l x¨ 1 = + cos θ θ˙ 2 + sin θ θ¨ 2 2 l x˙ 2 = − sin θ θ˙ 2 l l x¨ 2 = − cos θ θ˙ 2 − sin θ θ¨ 2 2 l y˙ 1 = cos θ θ˙ 2 l l y¨ 1 = − sin θ θ˙ 2 + cos θ θ¨ 2 2 y˙ 2 = l cos θ θ˙

(d)

y¨ 2 = −l sin θ θ˙ 2 + l cos θ θ¨ Since the system is scleronomic, energy is conserved and d (V + T) = 0 dt or kx1 x˙ 1 − mg y˙ 1 − Mg y˙ 2 + m˙y1 y¨ 1 +

ml 2 θ˙ θ¨ + Mx˙ 2 x¨ 2 + My˙ 2 y¨ 2 = 0 12

On substituting from Equation d and canceling out obtained 

 
m M m 3M m + + + cos2 θ θ¨ − + 12 4 4 4 4  g m k + M cos θ = 0 + sin θ (1 − cos θ ) − 4 l 2

(e)

˙ the following equation of motion is θ,  3M sin θ cos θ θ˙ 2 4 (f)

154

Dynamics of structures

If the system is not in motion, θ˙ and θ¨ are zero and Equation f will give the condition for static equilibrium (compare with Example 4.1).

4.10

HAMILTON’S EQUATION

In Section 4.9 it was pointed out that d’Alembert’s principle transforms the dynamical problem into a problem of static equilibrium, so that the method of virtual work can be applied to its solution. The resulting virtual work equation is given by Equation 4.67. In that equation the impressed forces Fai may be derivable from a scalar work function, but the inertia forces mi ri cannot be similarly derived. Hamilton showed that & by integrating the virtual work equation over time, the virtual work term mi r¨i · δri can be replaced by the variation of another scalar function: the kinetic energy of the system. Hamilton’s principle occupies an important position in analytical mechanics because it reduces the formulation of a dynamic problem to the variation of two scalar quantities: the work function and the kinetic energy, and because it is invariant under coordinate transformation. Multiplication of Equation 4.67 by dt and integration between limits t1 and t2 gives 

t2



(Fi − mi r¨i ) · δri dt = 0

(4.74)

t1

or 

t2



t1



t2

Fi · δri dt −



mi r¨i · δri dt = 0

(4.75)

t1

't The first integral on the left-hand side of Equation 4.75 can be written as t12 δW dt, where δW represents infinitesimal work. The second integral is reduced as follows 

t2

t1



mi r¨i · δri dt =



(t2  ( mi r˙i · δri ( − t1

t2 t1



mi r˙i ·

d (δri )dt dt

(4.76)

If it is assumed that the path of all particles is prescribed at time t1 and t2 , the variations δri ’s are zero at these times and the first term on the right-hand side of Equation 4.76 vanishes giving
  t2

 t2

dri mi r¨i · δri dt = − mi r˙i · δ dt dt t1 t1  t2

1 mi r˙i · r˙i dt δ =− 2 t1  t2

1 δ mi r2i dt =− 2 t1  t2 δT dt (4.77) =− t1

Principles of analytical mechanics 155

Substitution of Equation 4.77 into Equation 4.75 gives 

t2

δW + δT dt = 0

(4.78)

t1

Equation 4.77 is generally known as the extended Hamilton’s principle. In the special case when the impressed forces can be derived from a scalar work function, δW becomes a complete variation and Equation 4.78 becomes 

t2

δ(U + T)dt = 0

t1

or 

t2

δ

(U + T)dt = 0

(4.79)

t1

Recalling that the potential energy function V is the negative of work function, Equation 4.79 can be written as 

t2

δ

(T − V)dt = 0

(4.80a)

L dt = 0

(4.80b)

t1

or 

t2

δ t1

where L = T − V is called the Lagrangian. In a majority of engineering systems, the impressed forces are derivable from a potential energy function. The special form of Hamilton’s principle given by Equation 4.80 therefore applies to all such systems. Hamilton’s principle is a variational principle which states that of all possible paths, a mechanical system under motion will take that path which makes the integral in Equation 4.80 a minimum.

4.11

LAGRANGE’S EQUATION

Hamilton’s principle provides a complete formulation of a dynamical problem. However, for obtaining a solution to the problem, the Hamilton’s integral formulation should be converted into one or more differential equations of motion. These equations are of second order in the generalized coordinates and their number is equal to the number of degrees of freedom of the system. For holonomic constraints, the physical coordinates of the system can be expressed in terms of the generalized coordinates and the time variable as follows ri = fi (q1 , q2 , . . . , qN , t)

(4.81)

156

Dynamics of structures

Differentiation of Equation 4.81 gives N

∂fi ∂fi q˙ j + ∂qj ∂t

r˙i =

(4.82)

j=1

The kinetic energy T is given by

1

T=

2

i

mi r˙i · r˙i

(4.83)

Substitution of Equation 4.82 into Equation 4.83 indicates that the kinetic energy is a function of the following form T = T(q1 , q2 , q3 , . . . , qN , q˙ 1 , q˙ 2 , q˙ 3 , q˙ N , t)

(4.84)

Also, Equations 4.38 and 4.39 show that the infinitesimal virtual work is given by

δW =

N

Qj δqj

(4.38)

j=1

& where Qj = M i=1 Fi · ∂ri /∂qj . Using Equations 4.84 and 4.38, the Hamilton’s integral Equation 4.78 can be reduced to the following form by the methods of variational calculus N  t2



t1

j=1

d − dt




∂T ∂ q˙ j



 ∂T + + Qj δqj dt = 0 ∂qj

(4.85)

Since the virtual displacements δqj ’s are independent and arbitrary, it is possible to select a set of values in which δqj is zero for all values of j except j = m. Equation 4.85 will then reduce to 

t2



t1

d − dt




∂T ∂ q˙ m



 ∂T + + Qm δqm dt = 0 ∂qm

(4.86)

Again since δqm is arbitrary, the integral in Equation 4.86 can be zero only when the integrand is zero. The reasoning above leads to the following N Euler–Lagrange equations d dt




∂T ∂ q˙ j

 −

∂T − Qj = 0 ∂qj

for j = 1, 2, . . . , N

(4.87)

Principles of analytical mechanics 157

Next, consider the special case when the generalized forces can be derived from a potential energy function V, so that ∂V = −Qj ∂qj =−

M

j=1

Fi ·

∂ri ∂qj

(4.88)

It is apparent from Equations 4.81 and 4.88 that V will be of the form V = V(q1 , q2 , . . . , qN , t) The Lagranges equations are now obtained from the special form of Hamilton’s equation (Eq. 4.80) d dt




∂L ∂ q˙ j

 −

∂L =0 ∂qj

j = 1, 2, . . . , N

(4.89)

On setting L = T − V and noting that V does not depend on the velocities, q˙ j , Equation 4.89 can be written in the alternative form d dt




∂T ∂ q˙ j

 −

∂T ∂V + =0 ∂qj ∂qj

j = 1, 2, . . . , N

(4.90)

In general, a system will be subjected to some forces that can be derived from a scalar function and some others that cannot be so derived. In the latter class are forces of friction, air resistance, and some types of impressed forces. The Lagrange equation for such systems can be written as d dt




∂T ∂ q˙ j

 −

∂T ∂V + − Qj = 0 ∂qj ∂qj

j = 1, 2, . . . , N

(4.91)

where the Qj ’s now denote those forces that cannot be derived from a scalar function. Example 4.11 Obtain the equation of motion for the single-degree-of-freedom system shown in Figure E4.11.

Figure E4.11 Motion of a simple single-degree-of-freedom system.

158

Dynamics of structures

Solution The kinetic energy, potential energy, and the infinitesimal work are given by the following expressions 1 mq˙ 2 2 1 V = kq2 2 T=

(a)

δW = (−cq˙ + p sin t) δq = Q δq The equation of motion is obtained by substituting Equation a into the Lagrange’s equation d dt




∂T ∂ q˙

 −

∂T ∂V + −Q=0 ∂q ∂q

mq¨ + cq˙ + kq − p sin t = 0

(b)

Example 4.12 Obtain the equation of motion for the simple pendulum shown in Figure E4.12, in which the inextensible string supporting the mass m is being pulled up at a constant velocity v.

Figure E4.12 Simple pendulum of varying length.

Solution Angle θ between the string and the vertical is chosen as the generalized coordinate. The kinetic energy is given by T=

2 1 m (l − vt)θ˙ 2

(a)

The potential energy is a function of θ and t, so that the total energy is not conserved. V = {l − (l − vt) cos θ } mg

(b)

Principles of analytical mechanics 159 Also, ∂T = m(l − vt)2 θ˙ ∂ θ˙ 
d ∂T = m(l − vt)2 θ¨ + 2m(l − vt)(−v)θ˙ dt ∂ θ˙

(c)

∂V = mg(l − vt) sin θ ∂θ Substitution of Equations c into the Lagrange’s equation gives m(l − vt)2 θ¨ − 2mv(l − vt)θ˙ + mg(l − vt)sin θ = 0 or (l − vt)θ¨ − 2vθ˙ + g sin θ = 0

(d)

Example 4.13 Three interconnected rigid links, each of length l, are suspended from the ceiling by a hinge and subjected to a horizontal force P sin t at the free end of the lowest bar. Find the equations of motion.

Solution The system has three degrees of freedom. The three angles θ1 , θ2 and θ3 shown in Figure E4.13 are used as the generalized coordinates. The kinetic energy of each bar is composed of the energies due to the vertical and horizontal translations of the bar centroid, and the rotation of the bar

Figure E4.13 Motion of three interconnected links.

160

Dynamics of structures

about its centroid. The total kinetic energy of the system is given by T=

7 2 2 4 2 2 1 2 2 3 2 ml θ˙1 + ml θ˙2 + ml θ˙3 + ml θ˙2 θ˙1 cos(θ2 − θ1 ) 6 6 6 2 1 2 1 2 + ml θ˙3 θ˙1 cos(θ3 − θ1 ) + ml θ˙3 θ˙2 cos(θ3 − θ2 ) 2 2

(a)

The potential energy measured from a horizontal datum through the point of suspension is 5 3 1 V = − mgl cos θ1 − mgl cos θ2 − mgl cos θ3 2 2 2

(b)

The virtual work is given by   δW = δ P sin t{l sin θ1 + l sin θ2 + l sin θ3 } = Pl sin t [cos θ1 δθ1 + cos θ2 δθ2 + cos θ3 δθ3 ]

(c)

so that the generalized forces are Q1 = Pl sin t cos θ1 Q2 = Pl sin t cos θ2

(d)

Q3 = Pl sin t cos θ3 Substitution of Equations a, b, and d in Lagrange’s equations (Eq. 4.91) gives the following equations of motion 7 2 3 1 ml θ¨1 + ml 2 θ¨2 cos(θ2 − θ1 ) + ml 2 θ¨3 cos(θ3 − θ1 ) 3 2 2 1 3 − ml 2 θ˙22 sin(θ2 − θ1 ) − ml 2 θ˙32 sin(θ3 − θ1 ) 2 2 5 + mgl sin θ1 − Pl sin t cos θ1 = 0 2

(e)

3 1 4 2 ml θ¨2 + ml 2 θ¨1 cos(θ2 − θ1 ) + ml 2 θ¨3 cos(θ3 − θ2 ) 3 2 2 1 3 + ml 2 θ˙12 sin(θ2 − θ1 ) − ml 2 θ˙32 sin(θ3 − θ2 ) 2 2 3 + mgl sin θ2 − Pl sin t cos θ2 = 0 2

(f)

1 1 1 2 ml θ¨3 + ml 2 θ¨1 cos(θ3 − θ1 ) + ml 2 θ¨2 cos(θ3 − θ2 ) 3 2 2 1 1 + ml 2 θ˙12 sin(θ3 − θ1 ) + ml 2 θ˙22 sin(θ3 − θ2 ) 2 2 1 + mgl sin θ3 − Pl sin t cos θ3 = 0 2

(g)

Principles of analytical mechanics 161 Equations e, f, and g are highly nonlinear. If the system undergoes small vibrations, terms of second and higher order in θ can be neglected, and sin θ ≈ θ, cos θ ≈ 1. The resulting equations of motion are now linear and can be written in matrix form as follows  ml

2

   θ¨1 7/3 3/2 1/2 5/2 3/2 4/3 1/2   θ¨2  + mgl  1/2 1/2 1/3 θ¨3

   θ1 1   θ2  = Pl sin t  1  1 1/2 θ3 

3/2

(h)

Example 4.14 A uniform cylinder of mass m rolls without slip on a cart of mass M to which it is connected by a spring of stiffness k2 and a dashpot of coefficient c2 . The cart itself rolls on a horizontal surface and is connected to a wall by a spring of stiffness k1 and a dashpot of constant c1 . Find the equations of free vibrations of the system.

Figure E4.14 Cylinder rolling on a moving cart.

Solution As shown in Figure E4.14, the system has two degrees of freedom, the absolute translations of the cart and the cylinder. The angular velocity of the cylinder θ˙ is related to the two generalized coordinates as follows rθ˙ = q˙ 2 − q˙ 1

(a)

The kinetic and the potential energies of the system are given by T=

1 1 1 mr2 Mq˙ 21 + mq˙ 22 + 2 2 2 2

V=

1 1 k1 q21 + k2 (q2 − q1 )2 2 2




q˙ 2 − q˙ 1 r

2 (b)

(c)

The virtual work equation is δW = −c2 (q˙ 2 − q˙ 1 )δq2 + c2 (q˙ 2 − q˙ 1 )δq1 − c1 q˙ 1 δq1 = (c2 q˙ 2 − c2 q˙ 1 − c1 q˙ 1 )δq1 + (−c2 q˙ 2 + c2 q˙ 1 )δq2

(d)

162

Dynamics of structures

The generalized forces are thus Q1 = c2 (q˙ 2 − q˙ 1 ) − c1 q˙ 1 Q2 = −c2 (q˙ 2 − q˙ 1 )

(e)

Substitution of Equations b, c, and e into Lagrange’s equations gives the following equations of motion in the matrix form        M + m2 − m2 q˙ 1 k1 + k2 −k2 q1 0 q¨ 1 c1 + c2 −c2 + = + (f) 3m m −k2 k2 q2 q˙ 2 −c2 c2 q¨ 2 0 −2 2

4.12

CONSTRAINT COND ITIONS AND LAGRANG IAN MULTIPLIERS

At times it is more convenient to work with constrained coordinates rather than with generalized coordinates. When coordinates q are related through one or more constraint conditions, the variations δqj in Equation 4.85 are not independent and that equation does not lead to Lagrange’s equations (Eq. 4.87). In such situations, the method of Lagrange’s multipliers can be used effectively in the solution of the problem. The method is illustrated by the example that follows. Example 4.15 A circular cylinder of mass m and radius R1 rolls without slip on a fixed cylinder of radius R2 as shown in Figure E4.15. Find the equations of motion and the condition when the rolling cylinder will leave the surface of the fixed cylinder.

Figure E4.15 Cylinder rolling on a cylindrical surface.

Solution Let the motion be described by the two constrained coordinates, given by, θ, the angle that the line joining the centers of the cylinder makes with the vertical, and φ, the absolute rotation of

Principles of analytical mechanics 163 the rolling cylinder. These coordinates are related by the following constraint equation, which applies as long as contact persists between the two cylinders R2 θ − R1 (φ − θ ) = 0

(a)

The kinetic energy of the rolling cylinder is T=

1 1 mR21 2 ˙ (φ) m(R1 + R2 )2 (θ˙ )2 + 2 2 2

(b)

and the potential energy is V = A + mg(R1 + R2 )cos θ

(c)

where A is a constant. The equations of motion are obtained by minimizing the functional  t2 mR21 2 1 ˙ − A − mg(R1 + R2 )cos θ + λ [(R1 + R2 )θ − R1 φ] dt (d) (φ) m(R1 + R2 )2 (θ˙ )2 + 4 t1 2 The corresponding Lagrange’s equations are m(R1 + R2 )θ¨ − mg sin θ − λ = 0

(e)

1 mR21 φ¨ + λR1 = 0 2

(f)

Substituting for φ in terms of θ from Equation a and eliminating λ between Equations e and f, we get 3 (R1 + R2 )θ¨ − g sin θ = 0 2

(g)

which is the required equation of motion. Referring to Figure E4.15b, which shows the forces acting on the rolling cylinder, including those due to inertia, Equations e and f are easily recognized as the equations of equilibrium and λ can be interpreted as the force of friction at the contact between the two cylinders. Also, the normal reaction N that the stationary cylinder exerts on the rolling cylinder is given by ˙2 N = mg cos θ − m(R1 + R2 )(θ)

(h)

This force is positive as long as g cos θ > (R1 + R2 )θ˙ 2 , after which contact ceases.

4.13

LAGRANGE’S EQUATIONS FOR MULTI-DEGREE-OF-FREEDOM SYSTEMS

The equations of motion for a linear discrete multi-degree-of-freedom system can, of course, also be obtained by the applications of Lagrange’s equations. Consider, for example, an N degree-of-freedom system with generalized coordinates q1 , q2 , . . . , qN . The strain energy stored in the system is given by 1 T q fS 2 1 = qT Kq 2

V=

(4.92)

164

Dynamics of structures

where fS is the vector of forces that cause the displacements and K is the stiffness matrix. Equation 4.92 is called a quadratic form. Since the strain energy is always positive, the quadratic form of Equation 4.92 is always positive and is called a positive definite quadratic form. The kinetic energy is given by an expression similar to that for the strain energy but involving the generalized velocities instead of the displacements. Thus T=

1 T q˙ Mq˙ 2

(4.93)

Equation 4.93 is also a positive definite quadratic form. Differentiation of Equation 4.92 with respect to qj gives

1 ∂V = [0 ∂qj 2

  0 0    ..  1 T  .   0 · · · 1 · · · 0]Kq + q K    2 1 .   ..  0

(4.94)

where the row vector in the first term on the right-hand side has a 1 in column j with zeros elsewhere, while the column vector in the second term has a 1 in row j with zeros elsewhere. Because K is symmetric, the two terms on the right-hand side of Equation 4.94 are the transpose of each other, and since both are scalars, they must be equal. Equation 4.94 therefore becomes ∂V = [kj1 ∂qj

kj2 · · · kjN ]q

(4.95)

In a similar manner, ∂T = [mj1 ∂ q˙ j

mj2 · · · mjN ]q˙

(4.96)

and d ∂T = [mj1 dt ∂ q˙ j

mj2 · · · mjN ]q¨

(4.97)

Since ∂T/∂qj = 0, Lagrange’s equations become [ mj1

mj2 · · · mjN ]q¨ + [kj1

kj2 · · · kjN ]q = Qj

for j = 1, 2, . . . , N

(4.98a)

or Mq¨ + Kq = Q

(4.98b)

Principles of analytical mechanics 165

where Q is the vector of applied forces as well as the other nonconservative forces. Equations 4.98b represent the equations of motion of a linear discrete multi-degreeof-freedom system.

4.14

RAYLEIGH’S DISSIPATION FUNCTION

When a spring of stiffness k is stretched, the virtual work done by the spring force in moving through a distance δq is given by δW = −kqδq =


∂ ∂U 1 − kq2 δq = δq ∂q 2 ∂q

(4.99)

where U, the work function, is equal to −1 kq2 . As mentioned earlier, the negative 2 of the work done is designated as the potential energy function or the strain energy function of the spring, and the spring force is obtained from Fs = −

∂V = −kq ∂q

(4.100)

The dissipative forces arising in mechanical systems due to such sources as air resistance, internal friction, and acoustic vibrations are usually assumed to be proportional to the velocities along the physical coordinates and opposed to the motion. For a single coordinate, for example, the dissipative force or the damping force can be represented as arising from the motion of a dashpot in a viscous medium, as shown in Figure 4.14. If the dashpot constant is denoted by c, the damping force, Fd , is given by Fd = −cq˙

(4.101)

The virtual work done by the damping force is then ˙ δW = −cqδq

Figure 4.14 Analogy between spring force and damping force.

(4.102)

166

Dynamics of structures

Considering the analogy between the spring forces and the forces of damping, we can obtain a scalar function R of the generalized velocity q˙ from which the damping force Fd is obtained by an expression similar to Equation 4.100 Fd = −

∂R ∂ q˙

(4.103)

Function R is called the Rayleigh dissipation function; for the single dashpot shown, it is easily seen to be equal to 12 cq˙ 2 . In a general case the damping forces will be functions of more than one generalized velocity, and the dissipation function R should be such that δW =

N

−

i=1

∂R δqi ∂ q˙ i

(4.104)

Assuming that the generalized forces consist only of forces derived from a potential energy function and damping forces derived from a dissipation function as in Equation 4.104, Equation 4.87 becomes d dt




∂T ∂qj

 −

∂T ∂V ∂R + + =0 ∂qj ∂qj ∂ q˙ j

−

∂L ∂R + =0 ∂qj ∂ q˙ j

for j = 1, 2, . . . , N

or d dt




∂L ∂qj



(4.105)

Example 4.16 The mechanical system shown in the plan view in Figure E4.16 consists of three masses constrained to move between guides. Find the equations of motion.

Solution The motion of the system is described by the three generalized coordinates q1 , q2 , and q3 shown in Figure E4.16. The strain energy of a spring depends on the difference of the displacements parallel to the spring of its terminal points. The potential energy function of the system is thus given by V=

1 1 1 k1 q21 + k2 (q2 − q1 cos α)2 + k3 (q3 − q1 cos β)2 2 2 2

(a)

In an entirely analogous manner, the dissipation function is obtained as R=

1 1 1 c1 q˙ 21 + c2 (q˙ 2 − q˙ 1 cos α)2 + (q˙ 3 − q˙ 1 cos β)2 2 2 2

(b)

The kinetic energy of the system is T=

1 1 1 m1 q˙ 21 + m2 q˙ 22 + m3 q˙ 23 2 2 2

(c)

Principles of analytical mechanics 167

Figure E4.16 Three interconnected masses moving in frictionless guides.

Lagrange’s equations are therefore m1 q¨ 1 + k1 q1 − k2 (q2 − q1 cos α) cos α − k3 (q3 − q1 cos β) cos β +c1 q˙ 1 − c2 (q˙ 2 − q˙ 1 cos α) cos α − c3 (q˙ 3 − q˙ 1 cos β) cos β = 0 m2 q¨ 2 + k2 (q2 − q1 cos α) − c2 (q˙ 2 − q˙ 1 cos β) = 0

(d)

m3 q¨ 3 + k3 (q3 − q1 cos β) + c3 (q˙ 3 − q˙ 1 cos β) = 0 The equations above can be written in matrix form as follows Mq¨ + Cq˙ + Kq = 0  M=



m1



m2 m3

 k1 + k2 cos2 α + k3 cos2 β −k2 cos α −k3 cos β  −k2 cos α k2 0 K= −k3 cos β 0 k3   c1 + c2 cos2 α + c3 cos2 β −c2 cos α −c3 cos β  −c2 cos α c2 0 C= −c3 cos β 0 c3 

qT = [q1

q2

q3 ]

(e)

168

Dynamics of structures

SELECTED READINGS Fowles, G.R., Analytical Mechanics, 4th Edition, Saubders College Publishing, Philadelphia, Pa., 1986. Goldstein, H., Classical Mechanics, 2nd Edition, Addison-Wesley, Don Mills, On., 1980. Greenwood, D.T., Classical Dynamics, Prentice-Hall, Englewood Cliffs, 1977. Lanczos, C., The Variational Principles of Mechanics, 4th Edition, University of Toronto Press, 1970. Leipholtz, H.H.E., Six Lectures on Variational Principles in Structural Engineering, Solid Mechanics Division, University of Waterloo, 1978. Reddy, J.N., Energy and Variational Methods in Applied Mechanics, John Wiley, New York, 1984. Rosenberg, R.M., Analytical Dynamics of Discrete Systems, Plenum Press, New York, 1977.

PROBLEMS 4.1

A hollow cylinder of mass M, internal radius r1 and external radius r2 rolls without slipping inside a cylindrical surface of radius R (Fig. P4.1). By applying Lagrange’s equation, obtain the equation of motion for small vibrations of the system about a position of equilibrium.

Figure P4.1 4.2

Obtain expressions for the potential and kinetic energies of the system shown in Figure P4.2. Using the energy expressions derived by you, formulate the equations of motion. Linearize these equations assuming small vibrations.

4.3

In an attempt to reduce the vibrations of a floor deck, a vibration absorber is suspended from midspan of the deck. The floor deck can be idealized as a simply supported uniform beam of span L, mass m ¯ per unit length and flexural rigidity EI. The vibration absorber consists of a spring of stiffness k and a mass M attached to the deck as shown in Figure P4.3. Taking u1 and u2 as the generalized coordinates and assuming that the vibration shape of the floor deck is u = u1 sin (πx/L), obtain the Lagrange’s equations of motion.

4.4

Solve Problem 3.1 by using energy expressions and Lagrange’s equations.

Principles of analytical mechanics 169

Figure P4.2

Figure P4.3

4.5

Obtain the equations of motion for the system shown in Figure P4.5 by using Hamilton’s principle.

Figure P4.5

4.6

A slider of mass 3 kg slides in vertical frictionless guides and is restrained by a spring of stiffness 10 kN/m as shown in Figure P4.6. The slider is attached by a rigid link of mass 1 kg and length 1 m to a circular disc of mass 5 kg which rolls without slipping on a horizontal plane. In the position of equilibrium, the rigid link makes an angle of 45◦ with the vertical. Obtain the equations of motion for small vibrations of the system about its equilibrium position.

170

Dynamics of structures

Figure P4.6 4.7

Obtain the equations of motion for the system shown in Figure P4.7 in terms of generalized coordinates u and θ . Linearize the equations assuming small vibrations.

Figure P4.7 4.8

The axial mass moment of inertia of a wheel-axle assembly of weight W is determined by performing an experiment in which the assembly is allowed to roll without slipping on a pair of curved rails of radius R as shown in Figure P4.8. The motion of the system

Figure P4.8

Principles of analytical mechanics 171 can be described in terms of the constrained coordinates θ and φ shown in the figure. Write the energy expressions in terms of the coordinates shown. Then using the concept of Lagrangian multiplier obtain the equations of motion. Can a physical meaning be assigned in this case to the Lagrangian multiplier. 4.9

The point of attachment A of the simple pendulum shown in Figure P4.9 slides in a vertical slot; its displacement is given by y = G sin t. Obtain the equation of motion for the pendulum mass assuming that the vibration amplitude is small.

Figure P4.9 4.10

A cylinder of mass M and radius R rolls without slipping on a horizontal surface (Fig. P4.10). It is restrained by a spring of stiffness k and a damper with coefficient c. A simple pendulum consisting of mass m and a light rigid bar is attached by a pin to the center of the cylinder. Obtain the equations of motion for the cylinder and the pendulum considering small vibrations and assuming that: (a) the pin connecting the cylinder and the pendulum is locked so that the swing of the pendulum is equal to the roll of the cylinder; (b) the pendulum is free to rotate about the pin.

Figure P4.10

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Part 2

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Chapter 5

Free-vibration response: Single-degree-of-freedom system

5.1

INTRODUCTION

The procedures used to formulate the equation of motion for a single-degree-offreedom system were described in Chapter 2. When damping is viscous, the equation of motion invariably takes the form mu¨ + cu˙ + ku = p

(5.1)

Provided that m, c, and k do not vary with time, Equation 5.1 represents a linear second-order differential equation. Its solution depends not only on the nature of applied force but also on initial conditions from which motion is started. In fact, even when p is zero, the system will undergo vibrations if it is given an initial displacement or an initial velocity or both. The response of a system that is not subjected to any external force but is excited by initial disturbances alone is called free-vibration response. It is important to study such a response, not only because practical situations may arise when vibrations are excited by initial disturbances, but also because, when response to an applied force is desired, the complete solution must include a free-vibration component. When p is zero, Equation 5.1 becomes a homogeneous second-order differential equation of motion. In this chapter we study the solution of such an equation. We begin by analyzing the free-vibration response of a system in which the damping resistance is absent or negligible. Although systems without damping resistance do not exist in practice, the analysis of their response provides an insight into the nature of damped vibrations. Following the presentation of undamped free-vibration response, we discuss the free-vibration response of a damped system. As described in Chapter 2, damping forces may be of several different kinds. The easiest to handle mathematically is viscous damping. Other types of damping are hysteretic damping and Coulomb damping. We discuss the analysis of free damped response under each of the three types of damping resistances mentioned above. 5.2

UNDAMPED FREE VIBRATION

When damping is absent, the equation of motion becomes mu¨ + ku = 0

(5.2)

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Dynamics of structures

A possible solution of Equation 5.2 is of the form u = Geλt

(5.3)

where G and λ are arbitrary constants to be determined. Substitution of Equation 5.3 in Equation 5.2 gives Gλ2 meλt + Gkeλt = 0

(5.4)

Equation 5.4 will be satisfied provided that G = 0, but as evident from Equation 5.3, this leads to u = 0 and no motion takes place. For motion to take place, G must be nonzero and can be canceled from Equation 5.4. On canceling eλt as well, we obtain the characteristic equation λ2 m + k = 0

(5.5a)

λ = ±iω

(5.5b)

or

 where ω = k/m. The general solution of Equation 5.2 now becomes u = G1 eiωt + G2 e−iωt

(5.6)

By using de Moivre’s theorem, Equation 5.6 can be written in the alternative form u = A cos ωt + B sin ωt

(5.7)

where A and B are arbitrary constants. To determine the arbitrary constants, we use the specified initial conditions of displacement and velocity at time t = 0. u = u0 u˙ = v0

 at t = 0

(5.8)

Substitution of the initial conditions gives A = u0 and B = v0 /ω, so that Equation 5.7 becomes v0 sin ωt ω = ρsin(ωt + φ)

u = u0 cos ωt +

(5.9a)

Free-vibration response: Single-degree-of-freedom system

177

where ) ρ=

u20 +

 v 2 0

ω (5.9b)

u0 ω tan φ = v0

Equation 5.9 has been plotted in Figure 5.1c. It shows an oscillatory motion which repeats itself. Specifically, on examining the response at two instances of time 2π /ω apart, namely t1 and (2π/ω) + t1 , we get u(t1 ) = ρ sin(ωt1 + φ)

(5.10a)

 
  2π 2π + t1 = ρ sin + t1 ω + φ u ω ω


= ρ sin(ωt1 + φ) = u(t1 )

(5.10b)

It is easily proved that the velocity u˙ is also the same at the two instances of time. This implies that the motion repeats itself after T = 2π /ω seconds. This interval of time is called the period of motion. It takes T seconds from the time the system passes through a certain configuration moving in a certain direction until it next passes through the same configuration moving in an identical direction. This phase of motion is referred to as one cycle of motion. For the simple block–spring system of Figure 5.1a, the motion from the time the block is, say, in its zero-displacement position and moving to the right until the next instant of time when it is again in that position and is moving to the right can be described as one cycle of motion. The system executes f cycles of motion in 1 s, where f , called the natural frequency, is given by f =

1 ω = hertz (Hz) T 2π

(5.11)

The maximum displacement of the system is equal to ρ and is called the amplitude of vibration. It remains constant with time. Angle φ in Equation 5.9a is called the phase angle. The motion represented by Equation 5.9a and shown in Figure 5.1 is a harmonic motion of the simplest type and is referred to as a simple harmonic motion.

5.2.1 Phase plane diagram For conceptual visualization of a simple harmonic motion, it is useful to imagine a vector of length ρ starting off at an angle φ from the horizontal time axis and rotating anticlockwise with a constant angular velocity ω rad/s as shown in Figure 5.1b. The projection of the rotating vector on the vertical axis at time t is equal to ρ sin(ωt + φ)

178

Dynamics of structures

Figure 5.1 Free undamped vibrations of a single-degree-of-freedom system.

and gives the displacement of the system at that time. It is readily seen that as the rotating vector goes through one complete round, the system completes one cycle of motion. The angular velocity of the rotating vector, ω, is called the circular frequency of the system and is related to T and to f as follows ω = 2π f =

2π T

(5.12)

The rotating vector representation provides a convenient graphical method of obtaining the free-vibration response. It can also be extended to obtain the response under the action of an applied force. As already pointed out, the displacement u of a freely-vibrating system is given by Equation 5.9a, while the velocity u˙ is obtained by differentiating Equation 5.9a. It is

Free-vibration response: Single-degree-of-freedom system

179

useful to define a velocity function defined by u/ω ˙ and having the units of displacement. An expression for the velocity function can be obtained by differentiating Equation 5.9a and dividing by ω. Thus u˙ = ρ cos(ωt + φ) ω

(5.13)

Equations 5.9a and 5.13 are parametric equations of a circle of radius ρ. They represent the circle described by the rotating vector. The projection of this vector on horizontal axis is equal to the velocity function u/ω ˙ and can be used to construct the velocity– time diagram shown in Figure 5.1d. As already noted, the vertical component of the rotating vector is equal to the displacement u and gives the displacement–time diagram shown in Figure 5.1c. The curve described by the rotating vector is also referred to as the phase plane diagram, in which u/ω ˙ is measured along the abscissa and u is measured along the ordinate. It can be used to obtain the response for different kinds of initial conditions, for impulses applied during a response era, for sudden support motion, and even for an arbitrary applied load. In the following paragraphs, we illustrate several different applications of the phase plane diagram. As a simple example of the use of phase plane diagram, consider the free vibration response after an initial displacement of u0 . In this case, the length of the rotating vector obtained from Equation 5.9b is  ρ = u0 and the phase angle φ = π/2. The vector rotates at an angular speed of ω = k/m rad/s. The phase plane diagram is shown in Figure 5.2b. The response can be obtained directly from the phase plane diagram as  π u = u0 sin ωt + = u0 cos ωt 2  u˙ π = u0 cos ωt + = −u0 sin ωt ω 2

(5.14)

Alternatively, the displacement versus time and velocity function versus time relationships can be constructed graphically as shown in Figure 5.2c and d, respectively. Next consider the case when the motion is caused by an impulse I applied at time t = 0. From Newton’s law of motion, we know that the application of an impulse I results in an increase of velocity by I/m. In the case under consideration, therefore, the motion starts with an initial velocity v0 = I/m. The length of the rotating vector is v0 /ω and φ = 0. The resulting phase plane diagram is as shown in Figure 5.3b. The response is obtained either by graphical construction or by the following mathematical expressions derived from the phase plane diagram u=

v0 sin ωt ω

u˙ v0 = cos ωt ω ω

(5.15)

The displacement and velocity responses are shown in Figure 5.3c and d, respectively.

180

Dynamics of structures

Figure 5.2 Free vibrations from initial displacement.

As another example, consider a system undergoing free vibrations from initial conditions u0 and v0 . After t1 seconds an impulse I1 is applied to the system. As a result, its velocity changes by v1 = I1 /m. It is required to obtain the response of the system. The phase plane diagram for*the system is shown in Figure 5.4b. The starting

length of the rotating vector ρ = u20 + (v0 /ω)2 and its initial position is established by scaling v0 /ω on the horizontal axis and u0 on the vertical axis. In t1 seconds the vector has rotated through an angle ωt1 radians. At this time impulse I1 causes the velocity to change by I1 /m. On the phase plane diagram, a horizontal line of length I1 /(mω) is drawn to the right from the tip of the current position of the rotating vector. The new position of the rotating vector as well as its new length is now obtained by joining the origin to the end of the horizontal line just drawn. The second era

Free-vibration response: Single-degree-of-freedom system

181

Figure 5.3 Free vibrations from an initial velocity.

of response begins from this new position of the vector. Since the frequency of the system is unchanged, the vector continues to rotate at the initial speed of ω radians per second. The response can be obtained from the phase plane diagram either by graphical means or in the form of mathematical expressions u = ρ1 sin(ωt + φ1 )

0 ≤ t ≤ t1

u = ρ2 sin{ω(t − t1 ) + φ2 }

t1 < t

(5.16)

where ρ1 , ρ2 , φ1 , and φ2 are as indicated in Figure 5.4b. The displacement and velocity responses are shown in Figure 5.4c and d, respectively.

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Dynamics of structures

Figure 5.4 Vibrations of a single-degree-of-freedom system subjected to an impulse.

As a final example, suppose that the simple single-degree-of-freedom system shown in Figure 5.5a starts vibrating due to a sudden displacement ug of the support to the right. It is evident that the resulting motion of the block is the same as if the block were vibrating about a new origin shifted a distance ug to the right after it was given an initial displacement of −ug toward that origin. The resulting phase plane diagram is shown in Figure 5.5b, from which the response is seen to be u = ug (1 − cos ωt) u˙ = ug sin ωt ω

(5.17)

Free-vibration response: Single-degree-of-freedom system

183

Figure 5.5 Response to a sudden movement of support.

The displacement and velocity responses are shown in Figure 5.5c and d, respectively. Later we shall see that these responses are exactly equal to those caused by the sudden application of a constant force of magnitude kug . The potential energy of a system undergoing free vibration is equal to the strain energy stored in the spring, so that

V= =

1 2 ku 2 1 2 2 kρ sin (ωt + φ) 2

(5.18)

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Dynamics of structures

At the same time, the kinetic energy of the mass is given by 1 mu˙ 2 2

T= =

1 mω2 ρ 2 cos2 (ωt + φ) 2

=

1 2 kρ cos2 (ωt + φ) 2

(5.19)

Equations 5.18 and 5.19 give

V +T =

1 2 kρ 2

(5.20)

Since no energy is dissipated in a system undergoing free vibrations, the sum of the potential energy and the kinetic energy should be constant. This implies that as long as no external energy is input, the length of the rotating vector, ρ, should be a constant. Circles in a phase plane diagram thus represent constant-energy states. Application of external excitation such as an impulse or a support motion represent input of energy which causes a change in the length of the rotating vector. Example 5.1 With the aid of a phase plane diagram, obtain the displacement and velocity response of the simple single-degree-of-freedom system of Figure E5.1a to the support motion shown.

Solution The natural frequency of the system is given by ) ω=

k = m

)

50 = 10 rad/s 0.5

(a)

In the first era of response, the system vibrates about an origin located at 1.5 in. to the right. Since the rotating vector has a speed of 10 rad/s, in 0.1 s it will rotate through 1 rad or 57.3◦ as shown in Figure E5.1b. The phase plane diagram for the first era of motion is shown by arc AB in Figure E5.1b. The displacement response during this era is given by u = 1.5(1 − cos 10t)

(b)

and the velocity response is u˙ = 1.5 sin 10t ω

(c)

Free-vibration response: Single-degree-of-freedom system

185

Figure E5.1 Vibrations due to support motion.

At t = 0.1 s the displacement and velocity functions are u(1.0) = 0.69 in. u˙ (1.0) = 1.26 in. ω

(d)

The response in the second era of motion is represented in the phase plane diagram by the circle BCD with center at A. The length of the rotating vector in this second era is ρ=

 (0.69)2 + (1.26)2 = 1.438 in.

(e)

186

Dynamics of structures

while the new phase angle is φ1 = tan−1

0.69 = 0.501 rad = 28.7◦ 1.26

(f)

The displacement and velocity responses in the second era are given by u = 1.438 sin{10(t − 0.1) + 0.501} u˙ = 1.438 cos{10(t − 0.1) + 0.501} ω

(g)

The displacement–time and velocity–time relationships are shown in Figure E5.1c and d, respectively.

5.3

FREE VIBRATIONS WITH VISCOUS DAMPING

The equation governing the free vibration of a system with viscous damping is obtained from Equation 5.1 by setting p = 0. mu¨ + cu˙ + ku = 0

(5.21)

Equation 5.21 also has a solution of the form of Equation 5.3. Substitution of this solution in Equation 5.21 leads to the following characteristic equation in λ: mλ2 + cλ + k = 0

(5.22)

Solution of Equation 5.22 gives the following two values for λ: c λ1 = − + 2m c λ2 = − − 2m

) )

c2 k − 2 4m m

(5.23)

c2 k − 4m2 m

The characteristics of the damped free-vibration response will depend on the value of damping coefficient c. Three different cases arise 1 2 3

Critically damped system Overdamped system Underdamped system Each of the three cases is discussed in the following paragraphs.

5.3.1 Critically damped system A system in which the magnitude of damping is such that the discriminant in Equation 5.23 is zero is called a critically damped system. The damping constant c is, in this

Free-vibration response: Single-degree-of-freedom system

187

case, denoted by ccr and its value is given by √ c = ccr = 2 km = 2mω (5.24)  where ω = k/m is the natural circular frequency of the associated undamped system. The two roots of Equation 5.22 are now equal, so that ccr 2m = −ω

λ1 = λ2 = −

(5.25)

For repeated roots as in Equation 5.25, the general solution of Equation 5.21 is given by u = (G1 + G2 t)e−ωt

(5.26)

where G1 and G2 are arbitrary constants to be determined from initial conditions. Substitution of initial displacement and initial velocity from Equation 5.8 leads to the following values for G1 and G2 : G 1 = u0 G2 = v0 + ωu0

(5.27)

The general solution thus becomes u = {u0 + (v0 + ωu0 )t}e−ωt

(5.28)

Equation 5.28 has been plotted in Figure 5.6. It is noted that motion is not oscillatory. The system displacement decays exponentially with time and becomes very nearly zero after a while, although theoretically it takes an infinite time for the displacement to become zero. As we shall see later, critical damping is the least amount of damping for which the motion is nonoscillatory. Thus whenever damping is less than critical, motion becomes oscillatory.

Figure 5.6 Free-vibration response of a critically damped system.

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Dynamics of structures

5.3.2 Overdamped system When damping is greater than ccr , the system is said to be overdamped. For convenience, we define a damping ratio ξ given by ξ=

c ccr

(5.29)

For overdamped systems ξ > 1. Also, the damping constant c can be expressed as c = ccr ξ = 2mωξ

(5.30)

Substitution of Equation 5.30 in Equation 5.23 gives  λ1 = −ωξ + ω ξ 2 − 1  λ2 = −ωξ − ω ξ 2 − 1  If we denote ω ξ 2 − 1 by ω, ¯ the general solution of Equation 5.21 becomes ¯ ¯ u = e−ωξ t (G1 eωt + G2 e−ωt )

(5.31)

(5.32)

where the arbitrary constants G1 and G2 are again determined by initial conditions. Equation 5.32 can be expressed in the alternative form u = e−ωξ t (A cosh ωt ¯ + B sinh ωt) ¯

(5.33)

where A and B are also arbitrary constants. Equation 5.33 has been plotted in Figure 5.7. In this case, too, the motion is nonoscillatory. The displacement decays exponentially with time, although in comparison to a critically damped system, it takes longer for the system to return to a zero-displacement position.

Figure 5.7 Free-vibration response of an over damped system.

Free-vibration response: Single-degree-of-freedom system

189

5.3.3 Underdamped system In all structural systems and in a majority of mechanical systems, the damping is less than critical. Such systems are said to be underdamped. Mechanical systems for which it is required that the system return to a zero-displacement position in the least amount of time are designed to have critical damping. Examples are a recoiling gun and a weighing scale. Certain other recoil mechanisms, for example an automatic door closer, are designed to have overdamping. Leaving aside the few examples of the type cited above, most real systems are underdamped, and a study of underdamped vibrations is therefore of considerable importance. The damping ratio ξ is again defined as c/ccr , and the two solutions λ1 and λ2 take the form of Equation 5.31. The damping ratio ξ is however less than 1 in this case, the discriminant in Equation 5.31 is negative, and the two roots becomes imaginary. Thus  λ1 = −ωξ + iω 1 − ξ 2  λ2 = −ωξ − iω 1 − ξ 2

(5.34)

 Denoting ω 1 − ξ 2 by ωd , the general solution to Equation 5.21 can be written as u = e−ωξ t (G1 eiωd t + G2 e−iωd t )

(5.35)

By using de Moivre’s theorem, Equation 5.35 can be expressed in the alternative form u = e−ωξ t (A cos ωd t + B sin ωd t)

(5.36)

where A and B are arbitrary constants to be determined by the initial conditions. When A and B are determined from the conditions given in Equation 5.8, Equation 5.36 becomes u=e

−ωξ t




v0 + ωξ u0 u0 cos ωd t + sin ωd t ωd

 (5.37)

Equation 5.37 can be expressed as u = ρe−ωξ t sin(ωd t + φ)

(5.38)

where the amplitude ρ and the phase angle φ are given by 




ρ = (u0 ) + 2

u0 ωd tan φ = v0 + u0 ωξ

v0 + ωξ u0 ωd

2 1/2 (5.39)

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Dynamics of structures

Figure 5.8 Free vibrations of a damped system.

Equation 5.38 has been plotted in Figure 5.8c. The motion is oscillatory or cyclic and repeats itself after a period Td = 2π/ωd referred to as the damped period. Referring to the simple system of Figure 5.8a, a cycle of motion can be described as the motion taking place from the instant the block is at its extreme right position to the next instant of time, when it is again in that position. However, in this case the amplitude of displacement at the second instant of time is lower than that at the first. This amplitude decay is caused by the exponential term in Equation 5.38. The natural frequency of the system in cycles per second, denoted by fd , is given by fd = 1/Td . Figure 5.8c also shows curves representing two other functions of time: ρe−ωξ t and −ρe−ωξ t . These curves envelope the displacement–time relationship and touch the

Free-vibration response: Single-degree-of-freedom system

191

latter at those points where sin(ωd t + φ) is equal to +1 and −1, respectively. These points, however, do not represent maximas on the displacement–time relationship; the actual maximas lie just a bit to the left of these points. The time at which a maximum occurs is obtained by equating the derivative of Equation 5.38 to zero. The relationship giving the time at maximum displacement is found to be  tan(ωd t + φ) =

1 − ξ2 ξ

(5.40)

or  sin(ωd t + φ) = ± 1 − ξ 2

(5.41)

From Equation 5.41 it is evident that for small values of ξ , sin(ωd t + φ) is very nearly equal to ±1 at peaks in the displacement curve.

5.3.4 Phase plane diagram As in the case of undamped free-vibration response, a rotating vector representation can be used for damped free vibration, too. Imagine a vector of length ρe−ωξ t starting off at an angle φ from the time axis and rotating anticlockwise at a constant angular velocity of ωd radians per second. At any instant of time, the projection of such a vector on the vertical axis gives the displacement of the system at that instant. Unlike in the case of an undamped system, the size of the vector is not constant but decays exponentially with time. The speed of angular rotation, ωd , is called the damped circular frequency and it takes the vector one period = 2π /ωd seconds to complete one full rotation. The velocity expression for the system is obtained by differentiating Equation 5.38. Thus u˙ = −ωξρe−ωξ t sin(ωd t + φ) + ωd ρe−ωξ t cos(ωd t + φ)

(5.42)

Equation 5.42 can be written in the alternative form ωξ −ωξ t u˙ ρe sin(ωd t + φ) = ρe−ωξ t cos(ωd t + φ) − ωd ωd

(5.43)

Now suppose that we define an oblique pair of axes OX and OY, with axis OY inclined from the vertical by an angle σ as shown in Figure 5.8b. The component of the rotating vector of length ρe−ξ ωt on the vertical axis is ρe−ξ ωt sin(ωd t + φ). The component OB along the OX axis obtained by drawing AB parallel to the axis OY is OB = ρe−ξ ωt cos(ωd t + φ) − ρe−ξ ωt tan σ sin(ωd t + φ)

(5.44)

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Dynamics of structures

On comparing Equations 5.43 and 5.44, we note that OB will be equal to u/ω ˙ d, provided that we select σ such that tan σ =

ωξ ξ = ωd 1 − ξ2

(5.45)

Figure 5.8b represents the phase plane diagram for a damped single-degree-offreedom system. In this case, since the length of rotating vector decreases as e−ξ ωt , its trace is a spiral rather than a circle. The speed of the rotating vector is ωd instead of ω. As noted earlier, the projection of the rotating vector on the vertical axis is equal to the displacement and can be used to draw the displacement–time diagram of Figure 5.8c. The velocity function u/ω ˙ d is obtained by taking an oblique projection of OA on the horizontal axis. The oblique projection is obtained by drawing AB parallel to a line inclined at an angle σ from the vertical, where σ is given by Equation 5.45. The projected length OB can be used to construct the velocity function versus time relationship shown in Figure 5.8d. As in the case of an undamped system, the speed of angular rotation ωd is called circular frequency. It takes the vector one period, that is,  Td = 2π /ωd seconds, to complete one full rotation. The damped frequency ωd = ω 1 − ξ 2 is always less than the undamped frequency ω. However, for small values of ξ , the difference between the two is quite small, and in such cases ωd can be taken to be equal to ω without much error. As an example, for ξ = 0.1, ωd is 99.0% of ω. The construction of a phase plane diagram for a damped system involves somewhat more work than it does for an undamped system because the curve traced by the rotating vector is a spiral rather than a circle. The equation of the spiral r = ρe−ωξ t shows that the shape of the spiral depends on the damping ratio ξ . For a given value of ξ a spiral has to be drawn only once. Using this drawing as a template, the spiral can be transferred to the phase plane diagram by selecting the required value of ρ on the spiral. Thus, referring to Figure 5.9, if the starting length of vector is ρ = ρ1 , the spiral will begin from line OA, which is of length ρ1 , and the angle θ = ωd t will be measured from that line. At the end of time t the vector will be in position OB and its length will be given by OB = ρ1 e−ξ ωt = ρ1 e−ξ θω/ωd √ 2 = ρ1 e−θξ/ 1−ξ

(5.46)

For small values of ξ , OB = ρ1 e−θξ .

5.3.5 Logarithmic decrement In the free vibration of an underdamped system, displacement amplitude decays exponentially with time. The rate of decrease depends on the damping ratio ξ . If we denote the displacement at time t1 by u1 ≡ u(t1 ), then u(t1 ) = ρe−ξ ωt1 sin(ωd t1 + φ)

(5.47)

Free-vibration response: Single-degree-of-freedom system

193

√ 2 Figure 5.9 Spiral r = ρe−θξ/ 1−ξ .

The displacement at time t1 + 2π/ωd is given by
 2π u t1 + = ρe−ξ ω(t1 +2π /ωd ) sin(ωd t1 + φ) ωd

(5.48)

The ratio of u(t1 ) to u(t1 + 2π/ωd ) provides a measure of the decrease in displacement over one cycle of motion. This ratio is constant and does not vary with time; its natural log is called logarithmic decrement and is denoted by δ. The value of δ is given by  δ = ln

e−ξ ωt1



e−ξ ω(t1 +2π /ωd )

= 2π ξ

ω ξ = 2π  ωd 1 − ξ2

(5.49)

For small values of ξ , δ ≈ 2πξ . If δ is obtained from measurements and ξ is to be evaluated, we can use ξ=

δ 2π

(5.50)

or more accurately, ξ=√

δ 4π 2 + δ 2

(5.51)

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Dynamics of structures

Figure 5.10 Amplitude decay versus damping.

Over n cycles of motion, the displacement will decrease from u1 to un , the ratio of the two values being u1 = e2πξ nω/ωd un

(5.52)

Equation 5.52 gives ln

u1 = nδ un

(5.53)

The number of cycles N over which the displacement amplitude will decay to half its value at the beginning can be obtained from Equation 5.53 and is given by N=

ln 2 δ

(5.54)

The value of N obtained from Equation 5.54 is plotted in Figure 5.10 as a function of ξ . Example 5.2 The single-degree-of-freedom system with viscous damping shown in Figure E5.2a is displaced from its position of rest by a distance u0 and released. Obtain a plot of u/u0 versus ωt when the damping in the system is given by (a) ξ = 2, (b) ξ = 1, and (c) ξ = 0.1.

Free-vibration response: Single-degree-of-freedom system

195

Figure E5.2 Free-vibration response of a damped system with different levels of damping.

Solution (a)

For ξ = 2, the system is overdamped and the response is given by Equation 5.33: ¯ + B sinh ωt) ¯ u(t) = e−ωξ t (A cosh ωt

(a)

where  ω¯ = ω ξ 2 − 1 = 1.732ω The arbitrary constants A and B are evaluated by using the following initial conditions  At t = 0,

u = u0 u˙ = 0

The resulting values of A and B are A = u0 u0 ξ ω = 1.155u0 B= ω¯

(b)

196

Dynamics of structures Equation a now reduces to u = e−2ωt (cosh 1.732ωt + 1.155 sinh 1.732ωt) u0

(b)

Equation c is plotted in Figure E5.2b for ωt from 0 to 15. The response is nonoscillatory and decays exponentially with time. For ξ = 1, the system is critically damped and the response is given by Equation 5.28 with v0 = 0: u = (1 + ωt)e−ωt u0

(c)

(c)

(d)

Equation d has also been plotted in Figure E5.2b. Again, the response is nonoscillatory and decays exponentially with time. As compared to the overdamped system, the rate of decay is faster. For ξ = 0.1, the system is underdamped and will undergo cyclic motion. The response is given by Equation 5.37. Setting v0 = 0 in that equation, we get 
ωξ u0 sin ωd t u = e−ωξ t u0 cos ωd t + ωd

(e)

where  ωd = ω 1 − ξ 2 = 0.995ω Substituting for ξ and ωd in Equation e, we get u = e−0.1ωt (cos 0.995ωt + 0.1005 sin 0.995ωt) u0

(f)

Equation f is also plotted in Figure E5.2b for ωt from 0 to 15. As expected, the response is oscillatory and decays with time.

Example 5.3 For the system shown in Figure E5.2a, mass m = 91,000 kg and u0 = 30 mm. If the maximum displacement on the return swing is 20 mm at 0.5 s, determine (a) the spring stiffness k, (b) the damping ratio ξ , and (c) the damping constant c.

Solution Using the definition of logarithmic decrement, δ = ln

2πξ u1 =  u2 1 − ξ2

or ln

2πξ 30 =  20 1 − ξ2

(a)

Free-vibration response: Single-degree-of-freedom system

197

On solving Equation a for ξ , we get ξ = 0.0644

(b)

Also, Td =

2π = 0.5 s ωd

Hence  2π ωd = ω 1 − ξ 2 = 0.5

(c)

and ω=

2π = 12.593 rad/s  0.5 1 − ξ 2

k = mω2 = 91,000 × (12.593)2 = 1.443 × 107 N/m

(d)

The critical damping constant is obtained from ccr = 2mω = 2 × 91,000 × 12.593 = 2.292 × 106 Ns/m

(e)

The damping constant c is given by c = ξ ccr = 147,600 Ns/m

5.4

(f)

DAMPED FREE VIBRATION WITH HYSTERETIC DAMPING

As noted earlier, viscous damping force is given by cu, ˙ where c is a constant. Equation 5.42 shows that for small amounts of damping, velocity is proportional to ω. The damping force is thus seen to be proportional to the frequency of vibration and increases with the latter. As a result, the energy loss per cycle is also proportional to the frequency. Measurements of response, on the other hand, show that for most structural and mechanical systems, this is not true and the energy loss is either independent of frequency or in some cases decreases with increasing frequency. This is because in such systems a major part of damping occurs from internal friction, localized plastic deformation, or plastic flow, and strictly speaking, a viscous damping mechanism is not applicable. Damping resistance occurring from internal friction is referred to as hysteretic damping, structural damping, or solid damping. It results from the thermal effect of repeated elastic strain imposed on the material. Heat flowing across boundaries of grains causes dissipation of energy and hence damping of the motion. The loss of

198

Dynamics of structures

Figure 5.11 Hysteresis curve.

energy during repeated straining can be measured from a load–deformation curve of the type shown in Figure 5.11. The loop formed by the load–deformation curve is called a hysteresis loop and the area enclosed by it represents the loss of energy per cycle. The name “hysteretic damping’’ is thus derived from its relationship to the hysteresis loop. For a perfectly elastic material strained within its elastic limit the hysteresis loop degenerates into a straight line; consequently, the area enclosed by the loop and hence the energy loss becomes zero. In practice a perfectly elastic material does not exist and hysteresis energy loss is always present. It may be noted that when a material is strained beyond its elastic limit the stress-strain relationship becomes significantly nonlinear, and as explained in Section 2.4, the area enclosed by the hysteresis loop becomes quite substantial. The energy loss caused through global plastic deformation under stresses beyond the elastic limit is usually not modeled by a damping term in the equation of motion. Instead, a nonlinear force-displacement relation is used to determine the spring force fS . The resulting equation becomes nonlinear and a numerical method must be used for its solution. Methods that may be used in the solution of a nonlinear equation are briefly described in Chapter 20. For the present discussion we will assume that the material is not strained beyond its elastic limit and hysteresis energy loss can be accounted for by including a damping term in the equation of motion. If hysteretic damping is the only type of damping present in the system, the equation of motion takes the form ˙ =p mu¨ + fSt (u, u)

(5.55)

where p is zero for free vibrations. In a general case the solution of Equation 5.55 is quite complex. For harmonic motion, such as, for example, in the case of free vibrations with small amounts of damping, or, as we shall see later, in steady-state forced vibration under a harmonic excitation, hysteretic damping can be accounted for by assuming

Free-vibration response: Single-degree-of-freedom system

199

that in the equation of damped vibration, fSt consists of two components: fS = ku in which k is an average stiffness, and a damping force given by fD =

ηk u˙ 

(5.56)

where  is the frequency of vibration and η is a constant. Since for harmonic motion u˙ is proportional to the frequency of vibration, the latter cancels out from Equation 5.56, making the damping force independent of frequency. For free vibration,  ≈ ω, and if we define ch = ηk/ω, and ξh = ch /(2mω) = η/2, then because fD = ch u˙ from Equation 5.56, the free vibration equations (Eqs. 5.32 through 5.34) still apply, with c replaced by ch and ξ replaced by ξh . The free vibration response under hysteretic damping is thus given by u = ρe−ωξh t sin(ωd t + φ)

(5.57)

where * ωd = ω 1 − ξh2  2 1/2
+ ωξ u u 0 0 h ρ = u20 + ωd tan φ =

(5.58)

u0 ωd v0 + u0 ωξh

Also, the logarithmic decrement is obtained from δ ≈ 2π ξh = πη

5.5

(5.59)

DAMPED FREE VIBRATION WITH COULOMB DAMPING

Damping resistance may at times be provided by friction against sliding along a dry surface. The force of sliding friction, called Coulomb damping force, is proportional to the normal force acting on the contact surface, but is opposed to the direction of motion. Consider the simple block and spring system shown in Figure 5.12a. When the block is moving to the left (Fig. 5.12c), the friction is directed to the right and its magnitude is µN, where µ is the coefficient of friction. The equation of motion is given by mu¨ + ku = µN

(5.60)

When the block is moving toward the right (Fig. 5.12b), the friction force is directed toward the left and the equation of motion is mu¨ + ku = −µN

(5.61)

200

Dynamics of structures

Figure 5.12 Free vibration under Coulomb damping.

Equation 5.60 has a solution of the form

u = A cos ωt + B sin ωt +

µN k

(5.62)

where A and B are arbitrary constants to be determined from initial conditions. In a similar manner, Equation 5.61 has the solution

u = C cos ωt + D sin ωt −

µN k

(5.63)

where C and D are again arbitrary constants whose values depend on the initial conditions. To illustrate the essentials of free-vibration response under dry Coulomb friction, consider the case when the block has been displaced to the right a distance u0 , and released. During the first half cycle of motion that follows, the block is moving to the left and its motion is governed by Equation 5.62. To determine A and B, we substitute the initial conditions.  At t = 0,

u = u0 u˙ = 0

(5.64)

Free-vibration response: Single-degree-of-freedom system

201

Figure 5.13 Free vibration under Coulomb damping: (a) phase plane diagram; (b) displacement–time curve.

and obtain A = u0 −

µN k

(5.65)

B=0 Equation 5.62 now reduces to
 µN µN u = u0 − cos ωt + k k

(5.66)

Equation 5.66 is valid until the next instant of time, when the velocity becomes zero, that is, up to t = π/ω. At that instant the block is at its extreme left position and its displacement is −(u0 − 2µN/k). The block now starts moving to the right and its motion is governed by Equation 5.63 with the following conditions π At t = , ω



 u = − u0 − u˙ = 0

2µN k

 (5.67)

When C and D are determined by using the conditions in Equation 5.67, Equation 5.63 reduces to
 µN 3µN u = u0 − cos ωt − (5.68) k k Equation 5.68 is valid until the block again reaches its extreme right position, that is, when t = 2π /ω. At this instant the block has completed one cycle of motion and its displacement is u0 − (4µN/k). The complete motion of the block is described by the displacement–time curve of Figure 5.13b. The period of motion, that is, the time taken to complete one cycle of motion, is 2π/ω, which means that Coulomb friction

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Dynamics of structures

does not change the frequency or the period of vibration. The amplitude reduces by 4µN/k during each cycle of motion. At any instant when the velocity is zero, that is, when the block is either at its extreme left or extreme right position, if the displacement u is equal to or less than µN/k, the spring force ku will be equal to or smaller than the friction µN and the block will cease to move. Thus the block will come to rest at the end of a half-cycle in a position which is displaced from its original position of rest. In the foregoing discussion, we have not made any distinction between µs , the coefficient of static friction, and µd , the coefficient of dynamic friction. In general, µs is greater than µd . The dynamic coefficient of friction must be used in the equation of motion. However, the static friction is applicable in determining the position in which the block will come to rest. Thus at the end of any half-cycle when the spring force ku is less than the force of static friction µs N, the block will cease to move. The amplitude decay per cycle due to Coulomb friction is constant and is equal to 4µN/k. Thus the envelopes to the displacement–time curve are straight lines. It is of interest to note that with either viscous or hysteretic damping, amplitude decay is exponential, and theoretically the system never comes to rest.

5.5.1 Phase plane representation of vibrations under Coulomb damping A phase plane diagram can be used quite effectively to represent the vibrations of a system with Coulomb damping. Let us develop such a representation for the simple block shown in Figure 5.12. Let the block undergo vibrations after it has been displaced to the right a distance u0 and released. During the first half-cycle of its motion when the block is moving to the left, the motion is governed by Equation 5.66, which can be expressed as u−

µN = ρ1 sin(ωt + φ1 ) k

(5.69)

where ρ 1 = u0 −

µN k

π φ1 = 2

(5.70)

The velocity function is obtained by differentiating Equation 5.69, so that u˙ = ρ1 cos(ωt + φ1 ) ω

(5.71)

In the phase plane, Equations 5.69 and 5.71 represent the parametric equations of a circle as shown in Figure 5.13a. The circle has a radius ρ1 = O1 A and its center is located at (0, µN/k). The motion during the first half-cycle is represented by arc AB. The arc is drawn with O1 as the center and a radius of ρ1 .

Free-vibration response: Single-degree-of-freedom system

203

The second half-cycle of motion is governed by Equation 5.68, which can be expressed as u = ρ2 sin(ωt + φ2 ) −

µN k

(5.72)

where ρ 2 = u0 −

3µN k

π φ2 = 2

(5.73)

The velocity function is obtained by differentiating Equation 5.72 so that u˙ = ρ2 cos(ωt + φ2 ) ω

(5.74)

Equations 5.72 and 5.74 represent a circle with radius ρ2 and the center located at (0, −µN/k). In the phase plane diagram, the second half-cycle of motion is therefore represented by arc BC which is drawn with the center at O2 and a radius of O2 B. It can easily be shown that the third half-cycle of motion is represented by the arc CD drawn from center O1 with a radius O1 C. In summary, on a phase plane diagram the motion of a system with Coulomb damping is represented by arcs of circles with origins located at ±(µN/k) on the displacement axis. When the block is moving to the left, the center is on the positive direction of u axis, when it is moving to the right, the center is on the negative direction of u axis. The block comes to rest when at the end of a half-cycle of motion the absolute value of the displacement is less than µN/k. In the example of Figure 5.13, this happens at the instant represented by point F which lies within the region bounded by O1 O2 . Projection of the rotating vector on the vertical axis gives the displacement–time diagram (Fig. 5.13b). The velocity function versus time diagram is obtained from the projection of the vector on the horizontal axis. Example 5.4 The mass of 15 kg shown in Figure E5.4a is restrained by a spring of stiffness k = 1800 N/m and slides on a rough surface with a coefficient of friction µ = 0.1. The mass is given an initial displacement of 40 mm and an initial velocity of 250 mm/s and allowed to vibrate freely. At t = 0.25 s, the mass receives an impulse acting to the left of 2N s. At t = 0.5 s, the support suddenly moves to the right a distance of 30 mm. With the aid of a phase plane diagram, obtain the displacement–time relationship for the entire duration of motion of the block.

Solution The phase plane diagram is shown in Figure E5.4b. In free vibrations, the rotating vector will swing about origins located at ±µN/k = ±8.175 mm. The origins are indicated in Figure E5.4b by O2 and O1 . In the first era of vibration, the block is moving to the right; hence the rotating vector swings about O1 with an angular velocity given by ω=



k/m = 10.95 rad/s

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Dynamics of structures

Figure E5.4 Vibrations under Coulomb friction.

The initial position of the vector is obtained by locating the tip of the vector at coordinates u = 40 mm and u/ω ˙ = v0 /ω = 22.83 mm. The vector is shown by O1 A and its length is given by ρ1 =

 (40 + 8.175)2 + 22.832

= 53.31 mm The inclination of the rotating vector from the vertical, that is, angle AO1 B can either be calculated or measured from the diagram and is found to be 25.36◦ . In the second era of vibration, the block is moving to the left and the rotating vector swings about O2 . The length of the vector is ρ2 = O2 B = 36.96 mm. In 0.25 s, the vector should rotate through an angle of 0.25 × 10.95 rad = 156.85◦ . Angle BO2 C should therefore be 156.85 − 25.36 = 131.49◦ . At this instant, the block is still moving to the left. The impulse imparted to the block gives it an additional velocity of 2 × 1000/15 = 133.33 mm/s to the left. On the phase plane diagram, this is represented by the horizontal line CD of length 133.33/10.95 = 12.18 mm. The length of the vector now changes to O2 D, but it continues to swing about O2 until it reaches the extreme left position represented by point E. Angle O2 DE can be measured or calculated to be 58.60◦ . The total angle traversed by the vector up to this instant is 156.85 + 58.60 = 215.45◦ . Having reached the extreme left position represented by point E, the block will now start moving to the right, and the next era of motion is therefore represented by vector O1 E swinging about O1 . In 0.5 s the vector should swing a total of 0.5 × 10.95 = 313.69◦ . The angle of swing O1 EF is therefore 313.69 − 215.45 = 98.24◦ . At 0.5 s the support moves by a distance of 30 mm to the right. The origin on the phase plane diagram shifts an equal amount along the displacement axis from O to O . Simultaneously, O1 moves to O3 and O2 moves to O4 . The next phase of motion is represented by arc FG drawn

Free-vibration response: Single-degree-of-freedom system

205

with center at O3 . Following this phase, the rotating vector swings about O4 , and arc GH represents another half-cycle of motion. At H the block is in its extreme left position. The final half-cycle of motion is represented by arc HK drawn with center at O3 . Since K lies within the region O3 O4 , the force of friction is greater than the spring force and the block now comes to rest. The displacement–time relationship for the entire duration of motion is shown in Figure E5.4c.

SELECTED READINGS Crandall, S.H., “The role of Damping in Vibration Theory’’, Journal of Sound and Vibration, Vol. 11, 1970, pp. 3–18. Rao, S.S., Mechanical Vibrations, 3rd Edition, Addison-Wesley, Reading, Mass., 1995. Thomson, W.T., Theory of Vibration with Applications, 5th Edition, Prentice-Hall, Upper Saddle River, N.J., 1998.

PROBLEMS 5.1

A gun weighing 1500 lb is restrained by a spring and a damper (Fig. P5.1). The spring stiffness is selected so as to restrict the recoil of the gun to 4 ft. The damper is designed to engage only during the return to the firing position and the damping coefficient is adjusted so that the time taken for return to the firing position is a minimum. If the initial recoil velocity is 64 ft/s, determine the recoil spring stiffness and the damping required.

Figure P5.1

5.2

A machine weighs 200 lb and is supported on four springs and dampers such that the static deflection is 3/4 in. For vertical vibrations of the system, the dampers are adjusted to reduce oscillations to 1/4 of initial amplitude after two complete cycles. Find the damping coefficient and compare the frequency of damped and undamped oscillations.

5.3

A vehicle for landing on the moon has a mass of 4500 kg. The damped springundercarriage system of the vehicle has a stiffness of 450 kN/m and a damping ratio of 0.20. The rocket lift engines are cut off when the vehicle is hovering at an altitude of 10 m. Calculate the maximum deflection of the undercarriage system when the vehicle hits the surface. On the moon the gravitational acceleration is 1.6 m/s2 .

206 5.4

Dynamics of structures The packaging for a delicate instrument can be modeled as shown in Figure P5.4 in which the instrument of mass m is restrained by springs of total stiffness k inside a container of mass M. If the container is dropped on the ground through a height h and does not bounce on contact, obtain the maximum acceleration experienced by the instrument. Assume that during free fall the relative motion between masses M and m is negligible.

Figure P5.4 5.5

A wooden block of mass 3 kg is restrained by a spring of stiffness 2 N/mm (Fig. P5.5). A bullet of mass 0.2 kg is fired at a speed of 20 m/s into the block and embeds itself into the latter. Obtain the maximum displacement of the block, (a) neglecting damping; (b) assuming that damping is 10% of critical.

Figure P5.5 5.6

The single story frame shown in Figure P5.6 has a total story stiffness of 100 kips/in. The floor mass is 1 kip · s2 /in. The frame is subjected to a ground displacement as shown in the figure. Using a phase plane diagram, obtain the displacement history for the first 0.3 s. Neglect damping.

5.7

An automobile modeled as a single-degree-of-freedom system free to vibrate in the vertical direction. In a free-vibration test the amplitude of oscillation is observed to decrease from 20 mm to 2 mm in one cycle which takes 1/3 s. When carrying passengers with total mass 300 kg the free-vibration amplitude is observed to decay from 20 mm to 2.75 mm in one cycle. Find the mass of the vehicle and the stiffness and damping of the suspension system.

Free-vibration response: Single-degree-of-freedom system

207

Figure P5.6

5.8

A water meter consists of a float attached to a rigid light bar of length L (Fig. P5.8). The bar is attached to a wall by means of a hinge and is restrained by a damper of coefficient c attached at a distance a from the pin. The mass of the float is m, its cross-sectional area is A, and the mass density of water is ρ. If the damper is designed to provide critical damping, find its damping constant c. Neglect added mass of the water vibrating along with the float.

Figure P5.8

5.9

A mass M is supported on a firm base by a spring of stiffness k and a damper with damping constant c (Fig. P5.9). A mass m falls on the larger mass M from a height h and sticks to the latter without rebounding on impact. Find the equation governing the motion of the two masses after impact and determine their maximum displacement with reference to the position at impact. The following data are given: M = 25 kg, m = 5 kg, k = 1920 N/m, c = 48 Ns/m, and h = 1 m.

5.10

A 1 m long diving board with a diver of mass 75 kg standing at its tip oscillates with a frequency of 3 Hz. When the diver is standing still, the amplitude of oscillation is observed

208

Dynamics of structures

Figure P5.9

to decrease from 150 mm to 80 mm in 10 cycles. What is the modulus of rigidity EI for the board and what is the damping factor? Neglect the mass of the diving board. 5.11

The metacenter of a ship is defined as the point at which the resultant buoyant force intersects the centerline of the ship (Fig. P5.11). The distance from the center of gravity of the ship to the metacenter is referred to as the metacentric height. While in water the ship executes a rolling motion about the metacenter. The frequency of roll is observed to be ω. The mass of the ship is M and the mass moment of inertia for rotation about the center of gravity is J0 . Obtain an expression for the metacentric height assuming small vibrations.

Figure P5.11 5.12

An undamped system is observed to have a natural frequency of 20 rad/s. The system is equipped with a damper designed to provide critical damping. If it is now given an initial displacement, how much time will it take for the displacement to reduce to one-tenth its initial value.

5.13

A block of mass 3 kg is restrained by a spring of stiffness 2 N/mm and slides along a rough surface with a coefficient of friction of 0.2. An impulsive force applied to the block gives it an initial velocity of 1 m/s to the right. How far to the right will the block move? How

Free-vibration response: Single-degree-of-freedom system

209

much will be the maximum displacement of the block on its return swing towards the left? 5.14

How much time from the beginning of motion will it take for the block in Problem 5.13 to come to rest, and how much will be the offset between the starting point and the point at which the block comes to rest?

5.15

A single degree-of-freedom system of mass m and stiffness k is allowed to vibrate freely. Measurements show that over four cycles of motion, the amplitude of vibration reduces by a factor of 3.52. If the mass of the system is increase to 1.5m, by how much would the amplitude of vibration decrease over 4 cycles of motion when the damping in the system is (a) viscous; (b) hysteretic?

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Chapter 6

Forced harmonic vibrations: Single-degree-of-freedom system

6.1

INTRODUCTION

The vibrations in a dynamic system are usually caused by the presence of a time-varying force. The response that results from the application of such a force is called forced vibration response. When damping is absent or negligible, the vibrations are commonly referred to as undamped forced vibrations; when damping is taken into account, the term damped forced vibration is used. As discussed in Chapter 1, a system may be subjected to forces of several different types. When the applied force or forces are of a short duration, the resulting response is also of a comparatively short duration. After the applied force has ceased acting, damping in the system causes the vibrations to decay and the system returns to rest after a while. Response of this type is called transient response. Even though transient response is of a short duration, it may be quite significant from an engineering point of view. It is possible that certain components of the structural or mechanical system may be stressed beyond their capacity during the course of transient vibrations, resulting in a failure. In certain cases, failure may also result from a few repeated applications of a high stress level, a phenomenon called low cycle fatigue. Dynamic forces that act for a long duration cause the system to undergo sustained vibrations. Such vibrations have two components: a transient component and a steadystate component. The transient component is present at the beginning of the vibrations and its magnitude is governed by the initial conditions. Damping in the system causes it to decay rapidly. The steady-state component lasts as long as the exciting force. After the transient component has died, only the steady-state component remains. In certain situations, the steady-state response may build up to very large amplitudes, causing severe overstressing and possible failure. When such a condition occurs, the system is said to have achieved resonance. Even when resonance is not present, steady-state response may cause failure through fatigue. This is particularly significant for mechanical systems, which under steady vibrations could be subjected to a large number of stress cycles during their service life. There may also be cases where the total response in the initial phases consisting of the transient and steady-state components may be the most critical. An important class of long-term loads are those that can be expressed as harmonic functions of time. Such loads may, for example, arise due to imbalance in a rotating machine. Study of response under harmonic excitation, while useful by itself,

212

Dynamics of structures

also provides an insight into the nature of forced vibrations of a more general type. Therefore, after a brief introduction to the procedures used for solving the equations of motion under different types of loading situations, we discuss in this chapter the response to a harmonic excitation. In fact, any load that is periodic in nature can be treated by resolving it into its harmonic components. The response to each such component is obtained by methods presented in this chapter. These individual responses are then superimposed to obtain the total response. We defer the discussion of response to a periodic load to Chapter 9, where the general topic of analysis in frequency domain is discussed in detail. The analysis of transient response caused by short-term or impulsive loads as well as the response to a general dynamic loading are presented in Chapter 7.

6.2

PROCEDURES FOR THE SOLUTION OF THE FORCED VIBRATION EQUATION

When damping is viscous in nature, the equation of motion for a single-degree-offreedom system reduces to the form mu¨ + cu˙ + ku = p(t)

(6.1)

Provided that m, c, and k do not vary with time, Equation 6.1 represents a secondorder linear differential equation. Linearity is a useful property because it simplifies the solution of differential equations. The principle of superposition holds for linear equations. Thus, as an example, suppose that the applied force p(t) can be expressed as a sum of its components, say p1 (t) and p2 (t) p(t) = p1 (t) + p2 (t)

(6.2)

Also suppose that the solution of Equation 6.1 with p1 (t) as the exciting force is u1 (t) and that with p2 (t) as the exciting force is u2 (t). Then, if the principle of superposition holds, the response u(t) to an exciting force p(t) is given by u(t) = u1 (t) + u2 (t)

(6.3)

When Equation 6.1 is linear, its solution can be expressed as the sum of two parts: a complementary function and a particular integral. The complementary function is obtained by solving Equation 6.1 with p(t) = 0. The resulting equation is called a homogeneous equation; its solution was discussed in Chapter 5. As shown there, the solution of a homogeneous equation involves two arbitrary constants whose values are chosen to satisfy the initial conditions. The particular solution of Equation 6.1 depends on the exciting force, but does not involve any arbitrary constants. By itself, the particular solution will not satisfy the initial conditions of displacement and velocity at time t = 0. Complementary solutions are transient in nature because they decay with time as a result of the damping. On the other hand, a particular solution persists as long as the exciting force does and therefore represents the steady-state part of the solution.

Forced harmonic vibrations: Single-degree-of-freedom system

213

When the exciting force is a simple mathematical function, the particular solution can be obtained by a process of trial. In this chapter we use this procedure to obtain solutions for several different types of loads. Linear equations can also be solved by using the transform methods of mathematics. The two transform methods that can be most effectively used are (1) the Laplace transform, and (2) the Fourier transform. By taking a Laplace transform, the differential equation of motion can be converted to an algebraic equation in the transfer function. This equation can easily be solved to obtain the transfer function. Inverse transform of the solution then gives the desired response. Success of the Laplace transform method thus depends on the availability of the values of direct and inverse transforms, and the usefulness of the method is therefore limited to cases where the direct and inverse transforms are known or are easily determined. In the Fourier transform method, an arbitrary applied load is expressed as the sum of a series of harmonic components . Except in the case of a periodic load, the number of such components is theoretically infinite. The response to the component loads is then synthesized to obtain the resultant response in the time domain. The resolution and synthesis are accomplished by using direct and inverse Fourier transforms. The advantage of Fourier transform method lies in the fact that efficient numerical techniques are available for obtaining the required transforms. The Fourier transform method has therefore proved to be a very powerful tool in the response analysis of linear systems subjected to arbitrary loads. Details of the method are discussed in Chapter 9. Response to general dynamic loads can also be obtained by using the convolution theorem discussed in Chapter 7. When the convolution theorem is used, the solution process reduces to the evaluation of an integral, part of whose integrand is the forcing function. When the latter function is of a complicated nature, or when it is specified only in the form of numerical values at certain discrete intervals of time, numerical methods are used for the evaluation of the convolution integral. Some of these methods are discussed in Chapter 8. In addition to the procedures outlined in the foregoing paragraphs, it is also possible to obtain the response of a system to any arbitrary loading by a direct numerical integration technique. Such techniques are also discussed in detail in Chapter 8. Finally, it is useful to discuss the effect of the nature of damping on response as well as on the solution procedure. In a majority of structural and mechanical system, the level of damping is quite small; as a result, the response is oscillatory. The presence of damping ensures that oscillations will eventually die out. However, when damping is small, its short-term effect on response is negligible. In the case of transient response, it is the maximum level of response rather than its duration that is of importance, at least from an engineering point of view. Because such a maximum value will be attained during the initial stages, there would not be enough time for the damping forces to absorb any significant amount of energy from the system and the level of maximum response will not be affected appreciably by the presence of damping. Damping forces may therefore usually be neglected in the evaluation of response to impulsive forces. For steady-state vibration under long-term loads, the presence of damping does affect response. This effect is generally quite small except in the case of resonance when damping reduces the response very significantly. When damping is either hysteretic or of Coulomb type, the equation of motion becomes nonlinear and its solution is quite complex. For steady- state vibration under

214

Dynamics of structures

harmonic excitation, approximate solutions can be obtained for both hysteretic and Coulomb damping by defining an equivalent viscous damping in each case. In this chapter we discuss procedures for determining equivalent damping of this type.

6.3

UNDAMPED HARMONIC VIBRATION

Neglecting damping and assuming that the excitation is given by p(t) = p0 sin t, the equation of motion becomes mu¨ + ku = p0 sin t

(6.4)

The complementary solution to Equation 6.4 is the same as that given by Equation 5.7. For a particular solution, we try u = G sin t

(6.5)

where G is a constant chosen so as to satisfy Equation 6.4. Substitution into Equation 6.4 gives G(k − m2 ) sin t = p0 sin t

(6.6)

Equation 6.6 will be true for all values of t provided that G = po /(k − m2 ). The complete solution is now obtained by adding the complementary and particular solutions. Thus u = A cos ωt + B sin ωt +

p0 sin t k − m2

(6.7)

where, A and B are arbitrary constants to be determined such that the total solution given by Equation 6.7 satisfies the initial conditions. The complementary function represents the transient part of the solution, and for the slightest amount of damping resistance it dies out with time. This will become more evident when we consider damped harmonic response. If the initial displacement is u0 and the initial velocity is v0 , Equation 6.7 reduces to u = u0 cos ωt +

v0 p0 1 (sint − β sin ωt) sin ωt + ω k 1 − β2

(6.8)

where β is the frequency ratio given by β=

 ω

(6.9)

The transient part of the response is given by u1 = u0 cos ωt +

v0 p0 β sin ωt sin ωt − ω k 1 − β2

(6.10)

Forced harmonic vibrations: Single-degree-of-freedom system

215

Figure 6.1 Transient, steady-state, and total response of an undamped system to harmonic load; β = 0.5, u0 = 0, u˙ 0 = 0.

while the steady-state component of the displacement is u2 =

p0 1 sin t k 1 − β2

(6.11)

The static deflection δmax under a load p0 equal to the amplitude of harmonic excitation is obtained from δmax =

p0 k

(6.12)

The transient and steady-state displacements for a system with β = 0.5 and zero initial conditions are plotted in Figure 6.1 against t/T, where T is the natural period. Also shown is the total displacement, which is equal to the sum of the steady-state and transient displacements. In each case the displacement has been normalized by the static displacement δmax . It will be noted that the transient response is not zero even though the initial conditions are zero. In fact, the value of transient response is such that the total response obtained by a superposition of the transient response and the steady-state response satisfies the given initial conditions, which in this case are zero.

216

Dynamics of structures

Figure 6.2 Variation of response amplitude with frequency ratio.

We now define the ratio of dynamic deflection to the static deflection δmax as the dynamic load factor D. For steady-state response of an undamped system subjected to harmonic excitation, the dynamic load factor becomes D(t) =

1 sin t 1 − β2

(6.13)

The amplitude of dynamic load factor, Ad = 1/(1 − β 2 ), has been plotted in Figure 6.2 as a function of β, the frequency ratio. For small values of β, Ad  1 and the maximum displacement is very nearly equal to that produced under a static load of the same amplitude. The applied load in this case varies so gradually that the response is very nearly equal to that produced in a static case. When the frequency of the applied load is very high, that is, when β is very large, the displacement approaches zero. In this case the load is varying so rapidly that the system has no time to respond and remains at rest. For β = 1, the response is infinitely large. This condition is commonly referred to as resonance. It is clear that at or near resonance, that is, when the frequency of the applied force is equal to or near the natural frequency of the system, the resulting response may be catastrophic. In practical design, such a situation should of course be avoided. Further, we note that for  < ω, that is, for β < 1, Ad is positive and the system is said to be vibrating in phase with the applied force. In other words the applied force and the displacement are always in the same direction. For β > 1, Ad is negative and the system vibration is out of phase with the excitation, so that the displacement is always in an opposite direction to the applied force. To illustrate the nature of response, the static deflections are compared in Figure 6.3 with the steady-state dynamic deflections produced by a harmonic load

Forced harmonic vibrations: Single-degree-of-freedom system

217

Figure 6.3 Steady-state displacement response: (a) static displacement; (b) response for β = 0.5; (c) response for β = 2.

for two different values of β. The static deflection is obtained when ω is very large compared to , so that β ≈ 0, and Equation 6.11 gives δ=

p0 sin t k

(6.14)

Equation 6.14 is plotted in Figure 6.3a. The displacement–time relationship of Figure 6.3a can also be obtained from a rotating vector representation in which we imagine a vector of length p0 /k starting off from along the horizontal axis and rotating anticlockwise at an angular velocity of  rad/s. The projection of this vector on the vertical axis gives the static displacement at any instant of time.

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Dynamics of structures

In Figure 6.3b we show the steady-state displacement produced by a dynamic load with frequency ratio β = 0.5. The amplitude of displacement is, in this case, 1.33 times that in the static case. The dynamic displacement follows the applied load exactly so that both are in identical direction and both reach their peaks simultaneously. If a rotating vector representation is used, the response is seen to be equal to the projection on a vertical axis of a vector of length 1.33 p0 /k, starting off from along the horizontal axis and rotating anticlockwise with an angular velocity of  rad/s. On comparing with Figure 6.3a, we note that the force vector or static displacement vector and the steady-state dynamic displacement vector move along with each other and are therefore said to be in phase. Figure 6.3c shows the steady-state displacement under a dynamic load with β = 2. The amplitude of displacement is, in this case, one-third that of the static displacement and the direction of displacement is opposite to the direction of applied load. The rotating vector has a length of p0 /3k, and its starting position makes an angle of 180◦ from the horizontal axis. The displacement vector is thus 180◦ out of phase with the force vector. By following a procedure identical to that outlined in the foregoing paragraph, it can be shown that when the excitation is p(t) = p0 cos t, the steady-state response is given by u=

p0 1 cos t k 1 − β2

(6.15)

The rotating vector representation is still applicable except that the displacement is now equal to the projection of the rotating vector on the horizontal axis.

6.4

RESONANT RESPONSE OF AN UNDAMPED SYSTEM

It is of interest to examine, at resonance, the nature of the complete response given by Equation 6.8. We do this for the simple case when uo = vo = 0. With these initial conditions, Equation 6.8 reduces to u=

p0 1 (sin βωt − β sin ωt) k 1 − β2

(6.16)

Resonance occurs when β = 1. For this value of β, the numerator and the denominator in Equation 6.16 are both zero and the displacement becomes indeterminate. Its limiting values can, however, be determined by using L’Hospital’s rule. According to this rule both the numerator and the denominator are separately differentiated with respect to β. The limiting value of the resulting expression as β approaches 1 gives the required displacement. Thus p0 ωt cos βωt − sin ωt β→1 k −2β

lim u = lim

β→1

=

1 p0 (sin ωt − ωt cos ωt) 2 k

(6.17)

Forced harmonic vibrations: Single-degree-of-freedom system

219

Figure 6.4 Response of an undamped system at resonance (β = 1).

Equation 6.17 is plotted in Figure 6.4. The response is periodic with a period of 2π/ω. The amplitude of response continues to grow indefinitely. A measure of the rate of growth can be obtained by taking the difference of amplitudes at two successive peaks. The time at a peak is obtained by equating the differential of Equation 6.17 to zero, giving u˙ =

p0 2 ω t sin ωt = 0 2k

(6.18a)

or ωt = nπ

n = 0, 1, 2, . . .

(6.18b)

The difference between two successive peaks on the same side of the axis is then given by 
 nπ  nπ p0 2π + −u = − 2π cos nπ u ω ω ω 2k p0 =± π (6.19) k 6.5

DAMPED HARMONIC VIBRATION

When viscous damping is included, the equation of motion for a system subjected to excitation p(t) = p0 sin t becomes mu¨ + cu˙ + ku = p0 sin t

(6.20)

220

Dynamics of structures

The complementary part of the response is obtained by solving Equation 6.20 with the right-hand side equal to zero. For a damping value less than the critical, this solution is given by Equation 5.36. For the particular integral, we try a solution of the form u = G1 cos t + G2 sin t

(6.21)

where G1 and G2 are as yet unknown constants. Substitution of u and its derivatives from Equation 6.21 in Equation 6.20 gives (−G1 2 m + G2 c + G1 k) cos t + (−G2 2 m − G1 c + G2 k) sin t = p0 sin t

(6.22)

If Equation 6.21 is to be a valid solution, Equation 6.22 should hold for all values of t. In particular, it should hold when sin t is zero, so that cos t = ±1. This implies that the coefficient of cos t in Equation 6.22 should be zero. By similar reasoning we conclude that the coefficient of sin t in that equation should be equal to p0 . We thus obtain the following two equations in G1 and G2 −G1 2 m + G2 c + G1 k = 0 −G2 2 m − G1 c + G2 k = p0

(6.23)

On substituting c = 2ξ ωm and k = mω2 in Equations 6.23, dividing each side of the two equations by mω2 , and noting that /ω = β, we get (1 − β 2 )G1 + 2ξβG2 = 0 −2ξβG1 + (1 − β 2 )G2 =

p0 k

(6.24)

On solving Equation 6.24 for G1 and G2 , we obtain G1 =

p0 −2ξβ k (1 − β 2 )2 + (2ξβ)2

(1 − β 2 ) p0 G2 = k (1 − β 2 )2 + (2ξβ)2

(6.25)

The particular solution is now obtained from Equations 6.21 and 6.25 u=

1 p0 {(1 − β 2 ) sin t − 2ξ β cos t} 2 2 k (1 − β ) + (2ξβ)2

(6.26)

Equation 6.26 can be written in the alternative form u = ρ sin(t − φ)

(6.27)

Forced harmonic vibrations: Single-degree-of-freedom system

221

where ρ=

1 po  2 k (1 − β )2 + (2ξβ)2

tan φ =

2ξβ 1 − β2

(6.28a)

(6.28b)

 2 2 2 It is evident from the definition of angle φ that sin φ = 2ξβ/  (1 − β ) + (2ξβ) will always be positive. On the other hand, cos φ = (1 − β 2 )/ (1 − β 2 )2 + (2ξβ)2 will be positive when β < 1 and negative when β > 1. Angle φ should therefore lie between 0 and 180◦ . The complete solution is obtained by adding the complementary solution (Eq. 5.36) and the particular solution from Equation 6.26, so that u = e−ξ ωt {A cos ωd t + B sin ωd t} p0 1 + {(1 − β 2 ) sin t − 2ξβ cos t} k (1 − β 2 )2 + (2ξβ)2

(6.29)

where A and B are arbitrary constant determined such that the total solution given by Equation 6.29 satisfies the initial conditions. The complementary part of the solution represents the transient response which decays rapidly with time because of the exponential term e−ξ ωt . If the initial displacement is u0 and the initial velocity is v0 , the arbitrary constants A and B can be shown to be 2ξβ p0 + u0 k (1 − β 2 )2 + (2ξβ)2   v0 + u0 ωξ p0 ω 2βξ 2 − β(1 − β 2 ) + B= 2 2 2 k ωd (1 − β ) + (2ξβ) ωd

A=

(6.30a) (6.30b)

The transient part of the response thus becomes


 e−ωξ t v0 + u0 ωξ p0 u0 cos ωd t + sin ωd t + ut = e ωd k (1 − β 2 )2 + (2ξβ)2  

ω × 2ξβ cos ωd t + 2βξ 2 − β(1 − β 2 ) sin ωd t ωd −ξ ωt

(6.31)

while the steady-state part is given by Equation 6.26 or alternatively, by Equation 6.27. The transient and steady-state displacement for a system with β = 0.5, ξ = 0.1 and zero initial conditions are plotted in Figure 6.5. Also shown is the total displacement, which is equal to the sum of the steady-state and transient displacements. In each case the displacement has been normalized by the static displacement δmax . The transient response diminishes with time because of damping in the system. The total response obtained by a superposition of the transient response and the steady-state response satisfies the given initial conditions, which in this case are zero.

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Dynamics of structures

Figure 6.5 Transient, steady-state, and total response of a damped system to harmonic load; β = 0.5, ξ = 0.1, u0 = 0, u˙ 0 = 0.

As in the case of an undamped system, the dynamic load factor D(t) is defined as the ratio of dynamic deflection to the static deflection under load p0 . For steady-state response of a damped system, D(t) is given by D(t) = Ad sin(t − φ)

(6.32)

where Ad = 

1 (1 −

β 2 )2

+ (2ξβ)2

(6.33)

and φ is obtained from Equation 6.28b. Amplitude Ad is plotted in Figure 6.6 as a function of β for different values of ξ . A plot of the type of Figure 6.6 showing the variation of the amplitude of steady-state response with the exciting frequency is referred to as a frequency response function. As the exciting frequency approaches zero, the amplitude of steady-state displacement approaches the static displacement. ρ = p0 /k

(6.34)

Forced harmonic vibrations: Single-degree-of-freedom system

223

Figure 6.6 Variation of the amplitude of dynamic load factor with frequency ratio and viscous damping ratio.

Thus for small values of exciting frequency the response is controlled by the stiffness. For large values of the exciting frequency, that is large β, the limiting value of the displacement amplitude is given by ρ=

p0 p0 = 2 kβ m2

(6.35)

so that the response is controlled by the mass of the system. Of course, as  approaches infinity the displacement becomes zero. The value of β at which Ad is a maximum can be obtained by differentiating Equation 6.33 with respect to β and equating the differential to zero. This gives 1 −4β(1 − β 2 ) + 4β(2ξ 2 ) dAd =− =0 dβ 2 {(1 − β 2 )2 + (2ξβ)2 }3/2

(6.36a)

or β=

 1 − 2ξ 2

(6.36b)

Substitution of Equation 6.36b in Equation 6.33 gives the maximum value of Ad (AD )max =

1  2ξ 1 − ξ 2

(6.37)

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Dynamics of structures

Figure 6.7 Variation of phase angle with frequency ratio and damping.

Equation 6.37 shows that in the vicinity of a maximum the response is controlled by the damping. As in the case of an undamped system, resonance is said to occur when the exciting frequency is equal to the natural frequency, that is, when β = 1, and the exciting frequency  = ω is called the resonant frequency. Unlike the undamped case, however, maximum amplitude does not occur at resonance but at a value of β given by Equation 6.36b. This value of β is slightly less than 1. The corresponding value of the amplitude is (p0 /k)(1/2ξ 1 − ξ 2 ). When ξ is small, β for maximum amplitude is approximately equal to 1 and the maximum amplitude is (p0 /k)(1/2ξ ). It may be noted that in the literature some authors define resonant frequency as that frequency at which a response quantity, for example the displacement, is a maximum. As seen from Equation 6.36b the frequency at which the displacement is a maximum is slightly less that the frequency corresponding to β = 1, although for the amount of damping normally present in structures the two are not very different. In the present book we will continue to define a resonant frequency as the frequency corresponding to β = 1. Equation 6.27 shows that the displacement response of a damped system subjected to a harmonic load is itself harmonic, with a frequency equal to that of the exciting force but lags the latter by an angle φ called the phase angle. The phase angle, given by Equation 6.28b, is a function of both ξ and β. It is plotted in Figure 6.7 as a function of β for different values of ξ . At resonance, that is, when β = 1, φ is equal to 90◦ for all values of the damping ratio. This can also be verified from Equation 6.28b. To illustrate the nature of steady-state response, the static displacements are compared in Figure 6.8 with the steady-state displacements for three different values of β: 0.5, 1.0 and 2.0. and ξ = 0.1. For β = 0.5 the amplitude of displacement is 1.322p0 /k The phase angle is 0.133 rad or 7.6◦ . On the time axis this translates to a lag of (t/T) = 0.133/ T = 0.133/(β2π) = 0.042. For β = 1.0 the displacement amplitude is

Forced harmonic vibrations: Single-degree-of-freedom system

225

Figure 6.8 Steady-state response of a damped system with ξ = 0.1 to harmonic load: (a) β = 0.5; (b) β = 1.0; (c) β = 2.0.

5p0 /k. The phase angle is π/2 rad or 90◦ , which translates to a lag of t/T = 0.25 on the time axis. For β = 2 the displacement amplitude is 0.330p0 /k. The phase angle is 3.01 rad or 172.4◦ , which is equivalent to a lag of t/T = 0.24 on the time axis. The response of an undamped system can be obtained directly from that of a damped system by setting ξ = 0 in Equations 6.28a and 6.28b giving u=

1 po sin(t − φ) k |1 − β 2 |

(6.38)

where  φ=

0 π

for β < 1 for β > 1

Equation 6.38 is the same as Equation 6.11. It also shows that the response is in phase with the applied force when β < 1 and lags the applied force by 180◦ when β > 1. As in the case of an undamped system, a rotating vector can be used to represent the displacement response for a damped system, too. In Figure 6.9, the static displacement vector of length p0 /k is shown rotating counterclockwise at  rad/s. At any instant of time t, its inclination from the horizontal axis is t, and its projection on the vertical

226

Dynamics of structures

Figure 6.9 Rotating vector representation of steady-state displacements.

axis represents the static displacement at that time. The dynamic displacement vector of length ρ lags the static displacement vector by an angle φ. Its projection on the vertical axis is ρ sin(t − φ) and represents the dynamic displacement at time t. As shown in Figure 6.9, the dynamic displacement vector can be resolved into two components. One of these component vectors is in phase with the static displacement vector and hence with the applied force; its magnitude is ρ cos φ. The other component vector has a magnitude of ρ sin φ and lags the applied force vector by 90◦ . Figure 6.9 also applies to the case when the exciting force is p0 cos t, except that the displacement is now represented by the projection of the rotating vector on the horizontal axis. A rotating vector representation can also be used effectively to show the balance of forces acting on a system undergoing steady-state harmonic response. The force balance is expressed by the equation of motion, rewritten as p − f I − fD − fS = 0

(6.39a)

where p = p0 sin t fI = −m2 ρ sin(t − φ) fD = c ρ cos(t − φ)

(6.39b)

fS = kρ sin(t − φ) The force vectors are indicated in Figure 6.10. They all rotate in an anticlockwise direction at  rad/s. At time t, the applied force vector makes an angle of t from the horizontal axis. The spring force vector is of length kρ and is opposed to the displacement vector, which lags the applied force vector by angle φ. The projection of spring force vector on the vertical axis gives the magnitude of spring force at time t. The damping force vector is

Forced harmonic vibrations: Single-degree-of-freedom system

227

Figure 6.10 Force balance diagram of steady-state response.

of length cρ = 2ξ kβρ; it is opposed to the velocity vector and leads the spring force vector by 90◦ . Its projection on the vertical axis gives the damping force. The inertia force vector is of length m2 ρ = kβ 2 ρ and, as seen from Equation 6.39b, it is opposed to the spring force vector. Its projection on the vertical axis gives the inertia force. The sum of the vectors of the inertia force, damping force, and spring force is a vector that is equal and opposite to the applied force vector. The force balance representation applies as well in the case when p = p0 cos t, except that all projections must be taken on the horizontal axis. When required, the steady-state velocity and acceleration response of a system subjected to harmonic load can be obtained by successive differentiation of Equation 6.32. The velocity response is obtained from u˙ = Ad β cos(t − φ) p0 ω/k

(6.40)

The amplitude of the velocity response is thus given by Av = βAd

(6.41)

In a similar manner the acceleration response is given by u¨ = −Ad β 2 sin(t − φ) p0 ω2 /k

(6.42)

and the amplitude of acceleration is Aa = β 2 Ad

(6.43)

It is easily shown that  Av becomes a maximum at β = 1, while Aa attains its maximum value at β = 1/ 1 − 2ξ 2 . The maximum values of the velocity and acceleration amplitudes are (Av )max =

1 1 (Aa )max =  2ξ 2ξ 1 − ξ 2

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Dynamics of structures

Figure 6.11 Amplitudes of displacement, velocity, and acceleration response to a harmonic load for different damping ratios.

The displacement, velocity, and acceleration amplitudes are plotted as functions of β in Figure 6.11 for several different values of ξ . Because of the relationship among the three response quantities given by Equations 6.41 and 6.43, all of them can be plotted on a single four-way logarithmic graph as shown in Figure 6.12. In such a plot log Av is plotted on the ordinate against log β on the abscissae. Equation 6.41 gives log Av − log β = log Ad

(6.44)

For a constant Ad Equation 6.44 represents a straight line on the four-way log graph sloping to the right with a slope of 1. Lines sloping to the right having a slope of 1

Forced harmonic vibrations: Single-degree-of-freedom system

229

Figure 6.12 Four-way logarithmic graph showing response amplitudes for a system subjected to harmonic load.

thus represent constant Ad lines. A line perpendicular to these lines provides the log scale for displacement. In a similar manner Equation 6.43 gives log Av + log β = log Aa

(6.45)

For a constant Aa Equation 6.45 represents a straight line on the four-way log graph sloping to the left with a slope of −1. Lines sloping to the left with a slope of −1 thus represent constant Aa lines and a line perpendicular to these lines provides the log scale for acceleration. The four-way logarithmic graph in Figure 6.12 shows the variation of displacement, velocity, and acceleration amplitudes with the frequency ratio β for several different values of ξ . Example 6.1 The seating in a football stadium is mounted on precast prestressed concrete T-beams simply supported on a span of 11.7 m. The cross-sectional properties of a T-beam are indicated in 2 Figure E6.1. The superimposed load due to fixed seats and spectators can be taken as 2.4 kN/m . Clapping and stamping by spectators during a sporting event impress a harmonic dynamic load on the beams. Previous field observations have shown that the frequency of the harmonic 2 load due to stamping is 3 Hz, the load amplitude is 0.4 kN/m , and the damping is 3% of critical.

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Dynamics of structures

Figure E6.1 Dynamics of stadium seating: (a) cross section of precast concrete T-beam; (b) equivalent single-degree-of-freedom system. By assuming that the vibration shape function is   x 3  x 4  x ψ(x) = −2 + L L L convert a T-beam and the seating supported by it into an equivalent single-degree-of-freedom system. Then determine the natural frequency of the system, the maximum dynamic deflection at midspan, and the magnitude of the maximum acceleration experienced by the spectators.

Solution

The generalized mass m∗ , generalized stiffness k∗ , and the generalized load p∗ are given by  L 31 2 m∗ = mL ¯ m{ψ(x)} ¯ dx = 630 0  L 24 EI EI{ψ  (x)}2 dx = (a) k∗ = 5 L3 0  L ¯ pL ¯ pψ(x) dx = p∗ = 5 0 Mass per unit length, m ¯ Due to mass of the beam:

0.1587 × 2400 = 380.9 kg/m

Due to seats and spectators:

2400 × 0.762 = 186.4 kg/m 9.81

m = 567.3 kg/m 31 × 567.3 × 11.73 = 326.6 kg 630 24 × 3.7 × 107 × 6.327 × 10−3 k∗ = 5 × 11.73

m∗ =

= 701.6 kN/m

Forced harmonic vibrations: Single-degree-of-freedom system

231

The uniformly distributed harmonic load p¯ caused by stamping and clapping is obtained from p¯ = 0.40 × 0.762 sin t = 0.3048 sin t kN/m

(b)

where  = 2π × 3 rad/s. The effective force p∗ is thus given by p∗ = p0 sin t =

0.3048 × 11.7 sin t = 0.7137 sin t kN 5

The equivalent single-degree-of-freedom system is shown in Figure E6.1b. Its motion is governed by the following equations u = z(t) ψ(x)

(c)

m∗ z¨ (t) + c∗ z˙ (t) + k∗ z(t) = p∗ = p0 sin t

(d)

The solution of Equation d is obtained from Equation 6.27. Thus the displacement of the beam is given by u(x, t) = z(t)ψ(x) = ρ sin(t − φ)ψ(x)

(e)

where ρ=

p0 [(1 − β 2 )2 + (2βξ )2 ]−1/2 k∗

(f)

The frequency ratio β is obtained from the exciting frequency and the natural frequency. )

Natural frequency:

Frequency ratio:

) k∗ 701.6 × 1000 ω= = m∗ 326.6 = 46.35 rad/s = 7.38 Hz β=

3  = = 0.4065 ω 7.38

Substitution for β and ξ in Equation f gives ρ=

p0 0.7137 × 1.197 = 1.218 × 10−3 m 1.197 = k∗ 701.6

The midspan deflection is obtained from Equation e with x = L/2, so that
u

   x 3  x 4  L x , t = ρ sin(t − φ) −2 + 2 L L L x=L/2 = 1.218 × 10−3 × 0.3125 sin(t − φ) = 3.805 × 10−4 sin(t − φ) m

(g)

The maximum deflection at midspan is 3.805 × 10−4 m. The maximum acceleration at midspan is equal to 2 umax = (6π)2 × 3.805 × 10−4 = 0.135 m/s2 = 1.38% g, where g = 9.81 m/s2 is the acceleration due to gravity.

232

Dynamics of structures

Experience has shown that during a sports event, the spectators will not have a feeling of discomfort provided that the maximum acceleration experienced by them is below 5% g. The design is satisfactory from this point of view. 2 If w kN/m is the equivalent static load that will produce a deflection of 3.805 × 10−4 , 5 w × 0.762L4 = 3.805 × 10−4 384 EI from which we obtain w = 1.197 × 0.4 = 0.4788 kN/m

2

The precast beams should therefore be designed to carry a total superimposed load of 2 2.4 + 0.479 = 2.88 kN/m in addition to their self-weight.

6.6

COMPLEX FREQUENCY RESPONSE

The notation of complex algebra can be conveniently used to express the response to a harmonic force. If the forcing function is of the form p = p0 eit = p0 (cos t + i sin t)

(6.46)

the real part of the solution will represent the response to a cosine function, while the imaginary part will represent the response to a sine function. The steady-state response to the exciting force given by Equation 6.46 has the same frequency as the exciting force and can be expressed as u = Ueit

(6.47)

where U is the complex amplitude, also referred to as the complex frequency response. To obtain the value of U, we substitute for displacement and its derivatives from Equation 6.47 into the equation of motion (Equation 6.1). This gives (−m2 + ic + k)Ueit = p0 eit

(6.48)

Equation 6.48 leads to the following value of U U=

p0 −m2 + ic + k

(6.49)

By using the relationships: c = 2ξ ωm, ω2 = k/m, and β = /ω, we can express Equation 6.49 in the alternative form U=

p0 k(1 − β 2 + 2iξβ)

(6.50)

The complex frequency response given by Equation 6.49 or 6.50 contains information about both the amplitude and the phase of the response. This can be seen by expressing

Forced harmonic vibrations: Single-degree-of-freedom system

233

U in the following alternative form 1 p0 [(1 − β 2 ) − i2ξβ] k (1 − β 2 )2 + (2ξβ)2

U=

= ρe−iφ

(6.51)

where ρ is the real amplitude and φ is the phase angle, given by ρ=

1 p0  k (1 − β 2 )2 + (2ξβ)2

tan φ =

2ξβ 1 − β2

0≤φ≤π

(6.52a)

(6.52b)

Substitution of Equation 6.51 in 6.47 gives the following expression for the displacement response u = ρei(t−φ) = ρ{cos(t − φ) + i sin(t − φ)}

(6.53)

On comparing Equations 6.46 and 6.53 we conclude that the response to p0 sin t is ρ sin(t − φ), while the response to p0 cos t is ρ cos(t − φ). The complex frequency displacement response per unit force, Rd , is generally referred to as the receptance and is given by Rd =

u ρ −iφ U = e = it p0 e p0 p0

(6.54)

Two other related output parameters are also used to define the response to a harmonic force. The velocity response per unit force, Rv , is called mobility and is given by Rv =

u˙ = iRd = Rd eiπ/2 p0 eit

(6.55)

The amplitude of Rv is thus  times the amplitude of Rd while its phase angle is φ − π2 . The acceleration response per unit force is called inertance and is given by Ra =

u¨ = −2 Rd = 2 Rd e−iπ p0 eit

(6.56)

The amplitude of Ra is thus 2 times that of Rd , while the phase angle is φ + π . It will be noted that parameters U, V, and A are all complex and possess an amplitude and a phase angle, both of which are functions of the exciting frequency. Graphical representations of the complex frequency response functions must include three different parameters: the exciting frequency, the amplitude, and the phase angle. One form of graphical presentation consists of a plot of amplitude versus frequency, as shown in Figures 6.11 and 6.12, and a plot of phase angle versus frequency as shown in Figure 6.7. In another form the imaginary part of a function is plotted against the real part of the same function. Plots of this type are called Nyquist plots. They do not explicitly show the value of the exciting frequency. However, each point

234

Dynamics of structures

Figure 6.13 Nyquist plot of receptance; ξ = 0.2.

on the Nyquist graph corresponds to a specific value of the exciting frequency, which can therefore be identified on the plot. Figure 6.13 shows a typical Nyquist plot of receptance in which the imaginary and real parts are both obtained from Equation 6.51. Real(Rd ) =

1 1 − β2 k (1 − β 2 )2 + (2ξβ)2

1 2ξβ Imag(Rd ) = − 2 k (1 − β )2 + (2ξβ)2

(6.57)

The Nyquist plot shown in Figure 6.13 is close to a circle. Data points on the graph have been evaluated at regular increments of β and have been identified on the plot. Points near resonance are well separated and clearly distinguishable. Away from the resonant frequency the data points are very close together. This property of a Nyquist plot by which regions close to resonance are enlarged is very useful in the experimental determination of vibration characteristics, such as the natural frequency and damping. Figure 6.10 can be interpreted as a force balance diagram for the steady-state response of a system subjected to a complex harmonic force. The projections of the rotating vectors on the vertical axis, labeled imaginary axis, represent the imaginary parts of the forces, while those on the horizontal axis represent the real parts. In this form the force balance diagram of Figure 6.10 is referred to as the Argand diagram. Example 6.2 The single-degree-of-freedom system shown in Figure E6.2a has a mass m = 4,500 kg, stiffness k = 450 kN/m, and damping constant c = 18.0 kN · m/s. It is subjected to a harmonic force

Figure E6.2 (a) Single-degree-of-freedom system subjected to a harmonic force; (b) amplitude of frequency response function versus exciting frequency; (c) Nyquist plot of receptance; (d) Nyquist plot of mobility; (e) Nyquist plot of inertance.

Figure E6.2 Continued.

Forced harmonic vibrations: Single-degree-of-freedom system

237

p0 sin t. Plot the frequency response function on a log-log graph for exciting frequencies ranging from 0.1 to five-times the natural frequency. Also, obtain the Nyquist plots for receptance, mobility and inertance.

Solution Natural frequency: Damping ratio: Frequency ratio:

  3 /450 = 10 rad/s ω = k/m √ = 450 × 10 √ ξ = c/(2 km) = 18/(2 450 × 4.5) = 0.2 β = /10

The amplitude of the frequency response function U/p0 obtained from Equation 6.52a is plotted on a log-log graph in Figure E6.2b. for  ranging from 1.0 to 50 rad/s. For very small values of , the frequency response should approach 1/k. On the log-log graph 1/k plots as a horizontal straight line. This line is also shown on the graph. For large values of  the response should approach 1/(m2 ), which can also be represented by a straight line as shown on the log-log graph. The Nyquist plots for receptance, mobility, and inertance are shown in Figures E6.2c, d, and e, respectively. The mobility graph is a perfect circle, the others are close to a circle. The real and imaginary parts of the receptance are obtained from Equation 6.57. Mobility and inertance are obtained from receptance using the relationships given in Equations 6.55 and 6.56.

6.7

RESONANT RESPONSE OF A DAMPED SYSTEM

The complete response of a damped system is equal to the sum of a steady-state part given by Equation 6.26 and a transient part given by Equation 6.31. At resonance, that is, when β = 1 and hence  = ω, the steady-state solution, us , becomes

us = −

p0 1 cos ωt k 2ξ

(6.58)

while the transient part of the response, ut , is obtained by setting β = 1 in Equation 6.31. Assuming that the initial displacement as well as the initial velocity is zero, Equation 6.31 gives p0 e−ωξ t ut = k 2ξ


ωξ sin ωd t cos ωd t + ωd

(6.59)

For small values of damping, the contribution of the second term inside the parentheses in Equation 6.59 is negligible. As well, ωd ≈ ω, and Equation 6.59 therefore reduces to ut =

p0 e−ωξ t cos ωt k 2ξ

(6.60)

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Dynamics of structures

Figure 6.14 Resonant response of a damped system: (a) transient component; (b) steady-state component; (c) total response.

The steady-state component of the response given by Equation 6.58 is plotted in Figure 6.14b, for a value of ξ = 0.1. The transient response for the same damping is obtained from Equation 6.59 and is plotted in Figure 6.14a. Superposition of the two responses gives the total response shown in Figure 6.14c. As would be expected, the transient term cancels the initial displacement and initial velocity values obtained from the steady-state part so that the net values of these quantities is zero. It is evident from Figure 6.14c that the response rapidly builds up to its maximum value of p0 /(2ξ k). For a low damping ratio, the total response is equal to the sum of

Forced harmonic vibrations: Single-degree-of-freedom system

239

Figure 6.15 Variation of response amplitude with duration of loading at resonance.

Equations 6.58 and 6.60. The amplitude of response is, in this case, given by ρ=

p0 1 −ωξ t (e − 1) k 2ξ

(6.61)

Equation 6.61 approximates the envelope curves shown by dashed lines in Figure 6.14c. Such envelope curves are plotted in Figure 6.15 for different values of ξ and show the rate of buildup of response.

6.8

ROTATING UNBALANCED FORCE

Imbalance in rotating machinery is a common source of harmonic excitation. As an example, consider the rotation of a motor shown in Figure 6.16. The motor is rotating at a constant angular speed of  rad/s. The imbalance in the motor is represented by a small mass m0 attached to it at a radial distance e from the center. The coordinates x and y of the unbalanced mass are related to angle θ shown in the figure as follows x = e cos θ

(6.62a)

y = e sin θ

(6.62b)

Also, θ = t

(6.63)

240

Dynamics of structures

Figure 6.16 Unbalanced rotating mass.

The acceleration of the mass in the x direction is obtained by twice differentiating Equation 6.62a with respect to time x¨ = −e2 cos t

(6.64)

In a similar manner, the acceleration in the y direction is given by y¨ = −e2 sin t

(6.65)

These accelerations produce inertia forces m0 e2 cos t and m0 e2 sin t as shown in the figure. The inertia forces must, in turn, be resisted by the support system of the rotating motor. The unbalanced mass thus exerts harmonic forces in both a horizontal and a vertical direction. The resultant of the two forces is seen to be equal to m0 e2 acting along the radius vector. It is the well-known centrifugal force that acts on a rotating body. If the system shown in Figure 6.16 is free to vibrate in the vertical direction, its steady-state motion will be governed by the following equation mu¨ + cu˙ + ku = m0 e2 sin t

(6.66)

where m is the total mass of the motor and its base, including the unbalanced mass m0 , k is the total spring stiffness, and c represents the damping in the system. The steady-state response is given by Equation 6.26, so that u = ρ sin(t − φ)

(6.67)

Forced harmonic vibrations: Single-degree-of-freedom system

241

Figure 6.17 Variation of amplitude with frequency ratio and damping in an unbalanced rotating system.

where m0 e2 ρ=  k (1 − β 2 )2 + (2ξβ)2 =

eβ 2 m0 /m (1 − β 2 )2 + (2ξβ)2

tan φ =

2ξβ 1 − β2

(6.68a)

(6.68b)

In Equations 6.68, frequency ratio β = /ω,  and ω is the natural frequency of the motor and its support system given by ω = k/m. By equating the derivative of Equation 6.68a to zero, it can be shown that  amplitude ρ is a maximum when β = 1/ 1 − 2ξ 2 , and the maximum value is given by ρmax =

e(m0 /m)  2ξ 1 − ξ 2

(6.69)

For small values of ξ , the amplitude will be a maximum when β = 1 and ρmax = (em0 /m)(1/2ξ ). The dimensionless ratio ρ/(em0 /m) has been plotted in Figure 6.17 as a function of β for different values of the damping ratio ξ . It is observed that as would be expected,

242

Dynamics of structures

peaks in the response curve are slightly to the right of β = 1. Also, when β is large, amplitude ρ approaches a value em0 /m. It is evident that to keep the amplitude down, resonance should be avoided and the ratio em0 /m should be kept small. For practical design situations, this will be achieved by ensuring that the frequency ratio is significantly larger than 1 and the mass ratio m0 /m is kept as small as possible. In practice, low mass ratio is often obtained by mounting the machine on either a heavy frame or a massive concrete base. The expression for the phase angle is the same as Equation 6.28b, which applies to the case of a forcing function given by p = p0 sin t. The variation of the phase angle with β is therefore similar to that shown in Figure 6.7. Example 6.3 A steel frame consisting of four legs, each a steel wide-flange section W200 × 27, supports a rigid steel table as shown in Figure E6.3a. A rotating motor with an unbalanced mass of 200 kg at an eccentricity of 50 mm is mounted on the table. If the total mass of the table and the motor is 2500 kg, find the range of speed of the motor over which the maximum flexural stress in the legs will exceed 100 MPa. Assume that the legs have a negligible mass, are fixed at the foundation as well as at the table, and neglect damping.

Solution The lateral stiffness of the four legs is given by 4 × 12EI 4 × 12 × 200,000 × 25.8 × 106 = 3 L (3000)3

k=

= 9173 × 103 N/m Natural frequency of the system is obtained from ) ω=

k = m

)

9173 × 103 = 60.57 rad/s 2500 = 9.64 Hz

If the lateral deflection of the table is , the maximum moment M at the base as well as at the top of each leg is given by M=

6EI L2

The flexural stress σ is obtained from σ =

M S

where S is the section modulus. Thus 6EI L2 S 6 × 200,000 × 25.8 × 106 ×  100 = (3000)2 × 249 × 103 σ =

Forced harmonic vibrations: Single-degree-of-freedom system

243

Figure E6.3 (a) Steel frame supporting a motor; (b) displacement response versus speed of motor. or  = 7.24 mm For the maximum stress in the leg to exceed 100 MPa, the maximum lateral deflection must exceed 7.24 mm; that is, ρ in Equation 6.68a must exceed 7.24.

244

Dynamics of structures From Equation 6.68a we obtain ρ=

em0 β2  m (1 − β 2 )2

For β < 1, ρ≡

em0 β 2 > 7.24 m 1 − β2

50 × 200 β 2 > 7.24 2500 1 − β 2 or β > 0.802 For β > 1, ρ≡

em0 β 2 > 7.24 m β2 − 1

or β < 1.495 The lower limit of motor speed is βω = 0.802 × 9.64 Hz = 464 rpm. The upper limit is βω = 1.495 × 9.64 Hz = 865 rpm. In the range 464 to 865 rpm, the deflection will be larger than 7.24 mm and the flexural stress in the leg will be larger than 100 MPa. The amplitude of response ρ=

em0 β2  m (1 − β 2 )2

is plotted in Figure E6.3b as a function of the motor speed. The figure indicates the range of values of β over which the flexural stress will be greater than 100 MPa.

6.9 TRANSMITTED MOTION DUE TO SUPPORT MOVEMENT A system mounted on a moving support will have some of the support motion transmitted to it. Often the design of such a system requires that the transmitted motion be minimized. In other situations, the requirement may be to obtain estimates of the support motion by taking measurements of the transmitted motion. In either case, a relationship must be established between the motion of the support and that of the supported system. For harmonic excitation, this is easily accomplished. Consider the simple single-degree-of-freedom system shown in Figure 6.18a. The system consists of a mass m mounted on a frame through a dashpot and a spring. Let the frame move in a vertical direction according to the equation ug = G sint

(6.70)

Forced harmonic vibrations: Single-degree-of-freedom system

245

Figure 6.18 Harmonic support excitation: (a) frame supporting a mass; (b) free-body diagram of supported mass.

Let u represent the vertical displacement of mass m relative to the frame, while ut represents the same displacement with respect to a fixed frame of reference. Then the following relationship holds between ut , u, and ug ut = u + u g

(6.71)

Referring to the free-body diagram shown in Figure 6.18b, the equation of motion for mass m is given by mu¨ t + cu˙ + ku = 0

(6.72a)

or mu¨ + cu˙ + ku = −mu¨ g = mG2 sin t

(6.72b)

On comparing Equations 6.20 and 6.72b the steady-state response of the mass in Figure 6.18 is seen to be given by Equation 6.26 or 6.27 with p0 replaced by mG2 , so that u = ρ sin(t − φ)

(6.73)

where ρ=

1 mG2  k (1 − β 2 )2 + (2ξβ)2

= G

tan φ =

β2 (1 − β 2 )2 + (2ξβ)2

2ξβ 1 − β2

(6.74a)

(6.74b)

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Dynamics of structures

Figure 6.19 Variation of the amplitude of transmitted motion with frequency ratio and damping.

If we are interested in determining the total displacement of mass m, we can use Equation 6.71 and get ut = G sin t +

Gβ 2 {(1 − β 2 ) sin t − 2ξβ cos t} (1 − β 2 )2 + (2ξβ)2

G {(1 − β 2 + 4ξ 2 β 2 ) sin t − 2ξβ 3 cos t} + (2ξβ)2 (1 − = χ sin(t − γ )

=

β 2 )2

(6.75)

where % χ =G

1 + (2ξβ)2 (1 − β 2 )2 + (2ξβ)2

(6.76a)

tan γ =

2ξβ 3 (1 − β 2 ) + 4ξ 2 β 2

(6.76b)

In Figure 6.19, amplitude ratio χ /G has been plotted as a function of β for several different values of ξ . The rotating vector representation can also be used quite effectively to obtain the total displacement ut . Referring to Figure 6.20, the support displacement is represented by a vector of length G rotating at  rad/s, so that at time t it makes an angle t with

Forced harmonic vibrations: Single-degree-of-freedom system

247

Figure 6.20 Rotating vector representation of the displacement due to support excitation.

the horizontal axis. The displacement of the supported system relative to the support has an amplitude ρ and lags vector G by an angle φ. The total displacement is obtained by taking the vector sum of G and ρ and is given by the trigonometric identity χ 2 = G2 + ρ 2 + 2Gρ cos φ

(6.77)

On substituting for ρ from Equation 6.74a and noting that cos φ = (1 − β 2 )/  (1 − β 2 )2 + (2ξβ)2 , Equation 6.77 reduces to the form of Equation 6.76a. Phase angle γ between G and χ is obtained by applying the cosine identity to triangle OPQ ρ 2 = G2 + χ 2 − 2Gχ cos γ

(6.78)

It is easily shown that Equation 6.78 will lead to the same value of γ as that given by Equation 6.76b. Alternatively, angle η between vectors χ and ρ can be obtained from the relationship G2 = χ 2 + ρ 2 − 2ρχ cos η

(6.79)

Substituting for χ and ρ in Equation 6.79, we get cos η = 

1 1 + (2ξβ)2

(6.80a)

and tan η = 2ξβ

(6.80b)

Angle γ is now obtained from γ =φ−η

(6.81)

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Dynamics of structures

Example 6.4 If it is assumed that the stiffness of the tires is much greater than that of the springs, an automobile suspension can be idealized by a single-degree-of-freedom system shown in Figure E6.4. As the tire rolls along a washboard road, a vertical sinusoidal motion is imparted to the contact point. As a result, the car body also undergoes vertical vibrations. Assuming that h = 2 in., L = 20 ft, the suspension heave frequency is 1.2 Hz and the damping ratio is 0.85, determine the amplitude of motion of the car body when the car speed is 20 mph. Neglect the weight of the wheel and the axle. What would the amplitude be if the damper became ineffective?

Solution The period T of the harmonic vertical motion of the tire is equal to the time it takes to roll a distance L. The speed of the tire, V, is V=

20 × 5280 = 29.33 ft/s 3600

Hence T=

20 L = = 0.682 s V 29.33

and the frequency f is given by f =

1 = 1.467 Hz T

The frequency ratio β = f /f0 = 1.467/1.2 = 1.22. By substituting for β, ξ , and G = h = 2 in. in Equation 6.76a, we obtain the amplitude of motion of the car body % χ =h

1 + (2 × 0.85 × 1.22)2 {1 − (1.22)2 }2 + (2 × 0.85 × 1.22)2

= 2.16 in.

Figure E6.4 Automobile suspension on a washboard road.

Forced harmonic vibrations: Single-degree-of-freedom system

249

With ξ = 0, the amplitude χ will work out to 4.10 in. The shock absorber is thus effective in cutting down the amplitude of motion considerably.

6.10 TRANSMISSIBILITY AND VIBRATION ISOLATION In practical design it is frequently required that the dynamic forces transmitted by a machine to its surroundings be minimized. Vibration isolation of this type is referred to as force isolation and is usually achieved by inserting suitable springs and damping devices between the machine and its foundation. An inverse problem of isolation exists when it is required that vibrations from the surroundings are not transmitted to a supported system. Such isolation may be required when a delicate instrument is mounted inside a moving body or close to a source of vibrations, for example, in a space shuttle, an airplane, a ship, or near a floor supporting reciprocating machinery. In all these situations, the isolation problem is that of motion rather than of force. The two problems are, however, identical in most respects and the isolating devices used in each case have similar characteristics. Let us first consider the problem of force isolation. Figure 6.21 shows a machine of mass m mounted on a foundation through a set of springs and dampers, the total spring stiffness being k, and the damping coefficient c. Let the dynamic force acting on the machine in a vertical direction be p0 sin t. The steady-state response of the machine is u = ρ sin(t − φ), where ρ and φ are given by Equations 6.28. Force is transmitted to the foundation through the spring and the damper. The total transmitted force F is given by F = fS + fD

(6.82)

On substituting for fS and fD from Equations 6.39b, we get F = kρ sin(t − φ) + cρ cos(t − φ) = χ sin(t − φ + η)

Figure 6.21 Transmitted force under harmonic excitation.

(6.83)

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Dynamics of structures

where χ=

 (kρ)2 + (cρ)2

tan η =

c k

(6.84a) (6.84b)

On substituting for ρ from Equation 6.28 and noting that c/k = 2ξ/ω, Equation 6.84 reduces to % 1 + (2ξβ)2 (6.85a) χ = p0 (1 − β 2 )2 + (2ξβ)2 tan η = 2ξβ

(6.85b)

The ratio χ /p0 , called the transmission ratio or transmissibility, and denoted by TR, provides a measure of the force transmitted to the foundation. A comparison of Equations 6.76a and 6.85a shows that Figure 6.19 should also represent the variation of transmission ratio with β for different values of the damping ratio ξ . It is evident from Figure 6.19 that if the force is to be smaller than the applied force p0 , the √ natural frequency ω should be selected so that the ratio β = /ω is greater than 2. Also, √ for β > 2, the transmission ratio decreases with damping, so that, theoretically, zero damping will give the smallest transmitted force. In practice, however, some damping should always be provided to ensure that during startup as the machine passes through the resonant frequency, the response is kept within reasonable limits. √ In the design of isolation systems, when the frequency ratio β is greater than 2, the damping would usually be kept quite small. When damping forces are negligible, the expression for transmission ratio is greatly simplified, giving TR ≡

1 χ = 2 p0 β −1

(6.86)

The force balance diagram can also be utilized effectively to obtain the transmitted force. Such a diagram is shown in Figure 6.22. The transmitted force χ is obtained by taking the vector sum of the spring force vector kρ and the damping force vector cρ. Since the last two vectors are perpendicular to each other, their sum is given by Equation 6.84a. Also, the angle η between the spring force vector and the transmitted force vector is readily seen to be given by Equation 6.84b. We next direct our attention to the isolation of vibratory motion. In this case we are interested in determining the absolute movement of the supported system caused by a motion of the support. When the support undergoes harmonic motion, the amplitude of transmitted movement is given by Equation 6.76a. The similarity of this equation to Equation 6.85a implies that the design for motion isolation is governed by the same considerations that apply to force isolation. A variety of materials and devices are used in practice for vibration isolation. Isolators may consist of coiled springs of steel or pads of natural rubber, cork, or felt. In general, these materials serve both as springs and dampers. Coiled springs have a very small amount of damping. On the other hand, organic materials such as

Forced harmonic vibrations: Single-degree-of-freedom system

251

Figure 6.22 Vector diagram for transmitted force under steady-state harmonic excitation.

rubber, cork, or felt provide significant damping resistances. Metal springs are thus quite satisfactory when the supports can be designed so that the ratio√of the operating frequency to the natural frequency of the system, β, is greater than 2. When this is not possible, metal springs are not satisfactory. Example 6.5 A machine of mass 100 kg is supported on steel springs that deflect 1.2 mm under the weight of the machine. At the operating speed of the motor of 3000 rpm, imbalance causes a maximum disturbing force of 360 N. What is the maximum force transmitted to the foundation if damping in the steel springs is negligible? If the steel springs were replaced by neoprene pads having the same stiffness but a damping ratio of 0.2, what would be the maximum transmitted force?

Solution Denoting by  the static deflection under the weight of the motor, we have =

mg k

(a)

The natural frequency is now given by ) ω= ) =

k = m

)

g 

9.81 1.2 × 10−3

= 90.4 rad/s Hence the frequency in hertz, f0 = 14.4. The operating speed of the motor is 50 Hz, and the frequency ratio is therefore given by β = 50/14.4 = 3.47.

252

Dynamics of structures The transmission ratio for zero damping is obtained from Equation 6.86 TR =

1 1 = = 0.09 β2 − 1 (3.47)2 − 1

maximum transmitted force = TR × 360 = 32.4 N When damping is significant, the transmitted ratio is obtained from Equation 6.85a. Substituting ξ = 0.20 and β = 3.47 gives % TR =

1 + (2ξβ)2 (1 − β 2 )2 + (2ξβ)2

= 0.154 transmitted force = 0.154 × 360 = 55.3 N

Example 6.6 A sensitive instrument that requires to be insulated from vibration is to be installed in a laboratory where a reciprocating machine is in use. The vibrations of the floor of the laboratory may be assumed to be a simple harmonic motion having a frequency in the range 1000 to 3000 cycles per minute. The instrument is to be mounted on a small platform and supported on four springs arranged to carry equal loads. The combined mass of the instrument and supporting table is 5 kg. Calculate a suitable value for the stiffness of each spring if the amplitude of transmitted vibrations is to be less than 15% of the floor vibrations over the given frequency range. Assume that the damping is negligible.

Solution The frequency ratio should be larger than is then given by TR =

√ 2 for TR to be less than 1. The transmission ratio

1 β2 − 1

The condition that TR be less than 0.15 leads to 1 1 +1= + 1 = 7.67 TR 0.15 √ Hence the natural frequency f0 < f / 7.67 = f /2.77. The governing value of f will be the lower limit of the range of exciting frequency, that is, equal to 1000 cycles per minute. Hence β2 >

1 1000 × 60 2.77 = 6.02 Hz

f0 =

The total stiffness is now obtained from ) k 1 = 6.02 2π m

Forced harmonic vibrations: Single-degree-of-freedom system

253

or k = (6.02 × 2π )2 × 5 = 7154 N/m The stiffness of each spring is 1/4 × 7154 = 1788 N/m.

6.11 VIBRATION MEASURING INSTRUMENTS Measurement of vibrations plays an important role in engineering practice. The quantity to be measured may be a displacement, an acceleration, or a stress. The applications are quite diverse: from measurements of machine vibrations to the time history record of ground motion during an earthquake. In all cases, however, the underlying principle in the design of the instrument is the same. The instrument would usually consist of a spring–mass–damper system mounted on a rigid frame which is attached to the surface whose motion is to be measured. The quantity measured is the relative displacement between the frame and the instrument mass, and depending on the details of design, this displacement can be related to the displacement or acceleration of the support. The details of the design of a vibration measuring instrument, also known as a seismic instrument, are quite intricate. Usually, the instrument displacement must be magnified in some way before it can be measured. Such magnification may be achieved through mechanical levers or by optical means. A suitable damping device must be provided, and they range from an oil dashpot to an electric coil moving in a magnetic field. In some cases a permanent record must be obtained of the vibration history. This is usually accomplished by exposing a photographic film to a light beam or by using a stylus to scribe on a waxed paper. Details of design also depend on the application. Instruments for seismological observations have different requirements from those used for earthquake engineering measurements. Measurement of mechanical vibrations has its own special needs. Our primary interest here is to discuss the underlying principle in the design of seismic instruments. Figure 6.23 presents a schematic diagram of such an instrument. The instrument mass m is supported in the rigid frame through a spring of stiffness k and a damper with damping coefficient c. The frame of the instrument is rigidly attached to the surface whose motion is to be measured and moves along with the latter. The motion of the mass m relative to its support frame is measured by a suitable device and is correlated to a support motion parameter as discussed in the following paragraphs.

6.11.1 Measurement of support acceleration If the support is undergoing a harmonic motion with acceleration u¨ g = A sin t, then as discussed in Section 6.9, the equation governing the displacement of mass m relative to the frame is mu¨ + cu˙ + ku = −mA sin t

(6.87)

254

Dynamics of structures

Figure 6.23 Schematic diagram of a seismic instrument.

The solution to Equation 6.87 is given by Equations 6.27 and 6.28 with p0 replaced by −mA. Ignoring the negative sign on the right hand-side of Equation 6.87, we get u = ρ sin(t − φ)

(6.88)

where ρ= tan φ =

1 mA  k (1 − β 2 )2 + (2ξβ)2

(6.89a)

2ξβ 1 − β2

(6.89b)

and β is the ratio of the frequency of support motion to the natural frequency, ω, of the instrument. If the instrument is to be designed to measure an input acceleration which may, in fact, have several harmonic components of different frequencies, the measured displacement u should be proportional to the input for all values of the input frequency. From Equations 6.88 and 6.89a we note that the constant of proportionality is given by ρ 1 1 = 2 2 A ω (1 − β )2 + (2ξβ)2

(6.90)

For a satisfactory instrument design ρω2 /A should not vary with β. Figure 6.6 shows the variation of ρω2 /A with β for different values of ξ . It is observed that for ξ = 0.7, ρω2 /A stays approximately constant at a value of 1 provided that β lies between about 0 and 0.6. This is borne out by the figures in Table 6.1. If ξ = 0, then an instrument calibrated to be correct at a very small input frequency will be in error by 56% at an input frequency that is 60% of the instrument frequency. However, if ξ = 0.7, the error in measurement at β = 0.6 is only about 5.3%.

Forced harmonic vibrations: Single-degree-of-freedom system

255

Table 6.1 Characteristics of the output of a seismic instrument designed to measure acceleration. ρω2 /A β

ξ =0

ξ = 0.7

φ/β ξ = 0.7

0.0 0.1 0.2 0.3 0.4 0.5 0.6

1.00 1.01 1.04 1.10 1.19 1.33 1.56

1.000 1.000 1.000 0.998 0.991 0.975 0.947

1.400 1.405 1.419 1.442 1.470 1.502 1.533

It is evident from the discussion above that when the motion to be measured has a maximum frequency component with frequency equal to , the instrument should be designed to have a natural frequency ω > /0.6 = 1.67 if the error in measurement is not to exceed about 5%. Since, as seen from Equation 6.90, the relative displacement u is inversely proportional to the square of ω, a high instrument frequency will result in very small instrument displacements, and the latter must therefore be substantially magnified for proper measurement. Equation 6.88 also shows that the measured output of the instrument will lag the input motion by phase angle φ, implying that with respect to the excitation, the output will be shifted on the time axis by an amount ts = φ/ . In terms of β and ξ , this shift is given by 2ξβ 1 tan−1  1 − β2 2ξβ 1 tan−1 = βω 1 − β2

ts =

(6.91)

If the motion being measured comprises a single harmonic, this shift on the time axis will be of no particular significance. However, when the input motion has several harmonic components, the time shift being a function of β is, in general, different for each component output. The resulting shape of the total output, which is a superposition of its components is thus completely distorted and the measured displacement is unable to predict the input motion. It is therefore a requirement that ts = φ/  not vary with β. Fortunately, for ξ = 0.7, φ/ is practically constant, or φ is a linear function of  and hence of β. This is evident from both Figure 6.7 and Table 6.1.

6.11.2 Measurement of support displacement In the preceding section we discussed the design of an instrument for measuring support acceleration. If the quantity to be measured is support displacement rather than support acceleration, the design should be suitably modified. The harmonic input motion is now given by Equation 6.70 and the output by Equations 6.73 and 6.74.

256

Dynamics of structures

The amplitude of the input motion is G and that of the output is ρ. The variation of amplitude ratio ρ/G with β and ξ is indicated by Figure 6.17 provided that the ordinate in that figure is interpreted as ρ/G. For a well-designed instrument, this ratio should not vary with β, but should remain constant. From Figure 6.17 it is seen that for large values of β the amplitude ratio remains approximately constant at 1 irrespective of the value of ξ . The requirement that β should be large implies that the natural frequency of the instrument ω should be kept small. This can be achieved by providing a flexible spring and/or a heavy mass, both of which make the instrument unwieldy. For zero damping, Equation 6.74b gives φ = 180◦ and Equation 6.73 reduces to u = −ρ sin t. This implies that although the measured motion is negative of the input motion, there is no shift along the time axis and the output motion comprised of any number of harmonic components will be reproduced correctly. Example 6.7 An accelerometer is designed to have a natural frequency of 30 rad/s and a damping ratio of 0.7. It is calibrated to read correctly the input acceleration at very small values of the exciting frequency. (a) (b)

What would be the percentage error in the instrument reading if the harmonic motion being measured has a frequency of 20 rad/s? The displacement input imparted to the frame of the instrument is given by ug = 500 sin 10t + 400 sin 20t. Plot the variation of the input acceleration with time. Obtain the curve of instrument reading versus time and compare it with the input.

Solution (a)

From Equation 6.89a, 1 ρω2 =  A (1 − β 2 )2 + (2ξβ)2 where A is the amplitude of input acceleration. For β = 0, ρω2 /A = 1. The instrument is calibrated so that its reading is equal to ω2 times ρ. This ensures that at low frequencies the instrument will read the amplitude of input acceleration correctly. For β = 20/30 = 0.67 and ξ = 0.7, ρω2 = 0.921 A

(b)

This implies that for an input frequency of 20 Hz, the instrument reading ρω2 is 92.1% of A or 7.9% too low. The input acceleration is given by u¨ g = 500 sin 10t + 400 sin 20t

(a)

The two harmonic components of the input acceleration are plotted in Figure E6.6a and b, respectively. The instrument reading for the input motion with a frequency of 10 rad/s is given by 1 a1 = 500 ×  sin(10t − φ1 ) (1 − β1 )2 + (2ξβ1 )2

Forced harmonic vibrations: Single-degree-of-freedom system

257

Figure E6.7 Comparison of instrument measurement with the input acceleration. where β1 = 10/30 = 1/3, ξ = 0.7, and tan φ1 = 2ξβ1 /(1 − β12 ). Substitution of the given values yields a1 = 0.996 × 500 sin(10t − 0.483)

(b)

In a similar manner, the instrument reading for the input motion with a frequency of 20 rad/s is given by a2 = 0.921 × 400 sin(20t − 1.034)

(c)

258

Dynamics of structures

Relationships given by Equations b and c are shown on Figure E6.7a and b by dashed lines. It is observed that the curve for a1 is shifted by 0.483/10 = 0.0483 s with respect to the input acceleration, while the curve for a2 is shifted by 1.034/20 = 0.0517 s. Finally, the total input acceleration is compared with the resultant instrument reading in Figure E6.7c. Because the shift for the two components is slightly different, the output signal is a bit distorted, but the distortion is fairly small.

6.12

ENERGY DISSIPATED IN VISCOUS DAMPING

The energy balance in a system with viscous damping undergoing steady-state harmonic motion is of considerable interest. Energy is input into the system by the applied force p = p0 sin t. Denoting the infinitesimal work done by p as the system moves through a small distance du by dWe , we can write dWe = p0 sin t du

(6.92)

Since u = ρ sin(t − φ), du = ρ cos(t − φ) dt, and the time taken by the system to undergo one cycle of motion is 2π/ , the energy input per cycle becomes 

2π/ 

We =

p0 ρ sin t cos(t − φ) dt

0

= p0 ρπ sin φ

(6.93)

The energy input into the system is dissipated in viscous damping. We can prove this directly by considering the work done by the damping force per cycle of motion. Under steady-state vibration, the damping force is given by fD = cρ cos(t − φ)

(6.94)

The work done by fD as the system undergoes a small displacement du = ρ cos(t − φ) dt is given by dWi = c2 ρ 2 cos2 (t − φ) dt

(6.95)

The energy dissipated per cycle is obtained by integrating Equation 6.95 from 0 to 2π/  seconds. 

2π/ 

Wi = c ρ

2 2

cos2 (t − φ) dt

0

= cπ ρ 2

(6.96)

From Equations 6.26 and 6.28 we note that sin φ =  =

2ξβ (1 − β 2 )2 + (2ξβ)2

2ξβk ρ p0

(6.97)

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259

Figure 6.24 Force-displacement relationship under harmonic motion: (a) inertia force versus u; (b) spring force versus u; (c) damping force versus u; (d) spring force plus damping force versus u.

Substitution of Equation 6.97 into Equation 6.93 gives We = 2ξβkρ 2 π

(6.98)

Since ξ = c/ccr , ccr = 2k/ω, and β = /ω, Equation 6.98 reduces to We = cπ ρ 2 = Wi

(6.99)

The energy balance represented by Equation 6.99 implies that the work done by the spring and inertial forces per cycle is zero. This can also be proved by noting that

260

Dynamics of structures

fS = kρ sin(t − φ) and fI = −m2 ρ sin(t − φ) and integrating the infinitesimal work terms fS du and fI du over one cycle. The following alternative method is probably more illustrative of the concepts involved. The inertia and spring forces are related to the displacement as follows fI = −m2 u

(6.100a)

fS = ku

(6.100b)

The inertia force versus displacement relation is plotted in Figure 6.24a. It consists of a single straight line as shown. Since the area enclosed by the inertia force versus displacement curve over one cycle of motion is zero, the corresponding work done should also be zero. In a similar manner, the spring force versus displacement diagram drawn in Figure 6.24b is a straight line and shows that the work done per cycle by the spring force is also zero. Using Equation 6.94, the relationship between the damping force and displacement can be expressed as follows


fD cρ

2


2 u + =1 ρ

(6.101)

Equation 6.101 is the equation of an ellipse which is shown in Figure 6.24c. The area of ellipse is equal to cπρ 2 and represents the energy dissipated per cycle in viscous damping. The transmitted force fD + fS is obtained by superposition of Figure 6.24b and c. The transmitted force versus u diagram shown in Figure 6.24d is an ellipse whose principal axes are inclined with respect to the coordinate axes. The area of the skew ellipse is, of course, equal to the area enclosed by the ellipse shown in Figure 6.24c. In a similar manner the applied force, p = fI + fD + fS , versus the displacement diagram is obtained by the superposition of Figures 24a, b, and c. The resulting curve is again a skewed ellipse whose area is equal to the area of ellipse shown in Figure 24c. The energy balance at resonance can be derived from the foregoing discussion. In this case because β = 1, the spring force exactly balances the inertia force. This will be evident by comparing Equation 6.100a and b. The damping force should now exactly balance the applied force and the applied force versus displacement diagram should be identical to the damping force versus displacement diagram. The same conclusions can be drawn by referring to the force balance at resonance, Figure 6.25. Since, in this case, the phase angle φ = π/2, the spring force balances the inertia force and the damping force balances the applied force. In addition, Equation 6.93 reduces to We = p0 ρπ

6.13

(6.102)

HYSTERETIC DAMPING

Equation 6.96 shows that in a system with viscous damping the energy loss per cycle is proportional to the square of the amplitude and directly to the frequency of motion.

Forced harmonic vibrations: Single-degree-of-freedom system

261

Figure 6.25 Force balance diagram at resonance.

For a given amplitude, therefore, the energy loss will be proportional to frequency and will increase with the latter. As stated in Section 5.4, practical observations do not corroborate this and the actual energy loss per cycle is, in fact, independent of the frequency of motion. This is because damping forces are not viscous in nature but arise from internal friction, localized plastic deformation, and plastic flow that are present even when the global stress level is within the elastic range. It should be noted that our intent here is not to model the energy dissipated through plastic deformations at a global level introduced by stresses beyond the elastic limit. Such energy dissipation is usually accounted for by using in the analysis a nonlinear stress-strain relationship for the structural material under consideration. Methods of nonlinear analysis are discussed in Chapter 20. Damping caused by internal forces and local plasticity is referred to as hysteretic damping or rate-independent damping. The equation of motion for forced vibrations under hysteretic damping can be expressed as ˙ = p(t) mu¨ + fSt (u, u)

(6.103)

In a general case, the solution of Equation 6.103 is quite complex. However, for steady-state harmonic motion, hysteretic damping can be accounted for by expressing ˙ as the sum of two components: a spring force fS = ku, the total spring force fSt (u, u) where k is the average stiffness and a damping force fD given by fD =

ηk u˙ 

(5.56)

in which η is a constant. The energy loss per cycle due to hysteretic damping is obtained from Equation 6.96 by replacing c with ηk/  Wi = ηkπρh2

(6.104)

262

Dynamics of structures

The energy loss is seen to be independent of frequency but proportional to the square of the amplitude. The total spring force fSt is given by fSt = ku +

ηk u˙ 

(6.105)

and the equation of motion becomes mu¨ +

ηk u˙ + ku = p0 sin t 

(6.106)

√ We now define ch = ηk/  and ξh = ch /(2 km) = η/(2β), where η is a constant while ch and ξh both vary with β. The steady-state solution to Equation 6.106 is given by Equations 6.28a and b with ξ replaced by ξh . Thus noting that 2ξh β = η, u = ρh sin(t − φh )

(6.107)

where p0 1  k (1 − β 2 )2 + η2 η tan φh = 1 − β2

ρh =

(6.108a) (6.108b)

Amplitude ratio ρh /(p0 /k) and phase angle φh have been plotted in Figure 6.26a and b, respectively, as functions of β for several different values of η. The general shape of the curves is similar to that for viscous damping. However, the amplitude ratio is a maximum at β = 1, unlike the case of viscous damping, where the maximum occurs for a value of β slightly less than 1. Also, for β = 0, the phase angle φh is given by φh = tan−1 η, while in the case of viscous damping φ = 0 for β = 0. In other words, with hysteretic damping, the response is never in phase with the exciting motion. We can obtain an equivalent viscous damping coefficient by equating the energy loss per cycle in viscous damping to that given by Equation 6.104 2ξeq βkπρ 2 = ηkπρ 2 ξeq =

η 2β

(6.109)

It is apparent from Equation 6.109 that the equivalent viscous damping varies with the exciting frequency, so that we cannot use a single value for it throughout the frequency range of interest. However, as an approximation we may use in our analysis the equivalent viscous damping derived for a frequency equal to the resonant frequency. Thus setting β = 1 in Equation 6.109 we get ξeq = (ξ )β=1 =

η 2

(6.110)

Forced harmonic vibrations: Single-degree-of-freedom system

263

Figure 6.26 (a) Variation of the amplitude ratio with frequency ratio and hysteretic damping ratio; (b) variation of phase angle with frequency ratio and hysteretic damping ratio.

264

Dynamics of structures

Figure 6.27 Spring force versus displacement in a system with hysteretic damping.

The amplitude and phase angle of the frequency response obtained with equivalent viscous damping given by Equation 6.110 are shown in Figures 6.26a and b by dashed lines. The match between the two sets of curves is exact at β = 1 because the equivalent viscous damping was derived for that value of β. Away from resonance the two curves depart from each other, although the difference is not too large. It is of interest to plot the relationship between the total spring force fSt and the displacement u. From Equation 6.107, u˙ = ρh  cos(t − φh ). Substituting this in Equation 6.105, we get fSt − ku = cos(t − φh ) ηkρh

(6.111)

Since u = ρh sin(t − φh ), Equation 6.111 gives


fSt − ku ηkρh

2


+

u ρh

2 =1

(6.112)

Equation 6.112 is plotted in Figure 6.27 with force fSt on the ordinate and displacement u on the abscissa. The resulting curve is a skewed ellipse as shown. The area enclosed by the curve represents the energy loss per cycle and its value is ηkπρh2 . Equation 6.112 does not contain the exciting frequency . This implies that the hysteresis loop represented by Equation 6.112 can be obtained experimentally by carrying out a quasi-static cyclic load test in which the rate of loading is fairly slow, and plotting the resulting load–displacement relationship. The relationship shown in Figure 6.27 can be interpreted as the hysteresis loop of the structure. Intercept d on the displacement axis is given by d = ρh 

η

(6.113)

1 + η2

from which the damping fraction η is obtained as η= *

d ρh2 − d 2

≈

d ρh

(6.114)

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265

For light damping, intercept d is small in comparison to ρh and the approximation in Equation 6.114 applies. 6.14

COMPLEX STIFFNESS

For simple harmonic motion, hysteretic damping can conveniently be expressed by using the concept of complex stiffness. Thus if the exciting force is p0 eit , the displacement and velocity are given by u = Ueit

(6.115a)

u˙ = Uieit

(6.115b)

and

The hysteretic damping force can now be expressed as fD =

ηk u˙ 

= Uiηkeit

(6.116)

= iηku By substituting Equation 6.116 into Equation 6.106, the equation of motion is expressed as mu¨ + (1 + iη)ku = p0 eit

(6.117a)

¯ = p0 eit mu¨ + ku

(6.117b)

or

The quantity k¯ = k(1 + iη) is called complex stiffness. It is useful in taking account of hysteretic damping, particularly when the analysis is carried out in the frequency domain as described in Chapter 9.

6.15

COULOMB DAMPING

When damping resistance in a system arises due to dry friction, the equation of motion for harmonic excitation becomes mu¨ + ku ± µN = p0 sin t

(6.118)

Referring to Figure 6.28a the positive sign for the friction force applies when the mass m is moving to the right while the negative sign applies when the mass is moving to the left. Equation 6.118 is a nonlinear equation, and in a general case, its solution is quite complex. However, for small amounts of damping, an approximate but simpler

266

Dynamics of structures

Figure 6.28 System with Coulomb friction subjected to harmonic excitation.

solution can be obtained for steady-state harmonic vibrations. From the friction force versus displacement diagram shown in Figure 6.28b, it is apparent that the work done by the friction force per cycle of motion is given by Wi = 4µNρ

(6.119)

We can obtain an equivalent viscous damping coefficient by equating the energy loss per cycle in viscous damping to that given by Equation 6.119 ρ 2 π ceq = 4µNρ ceq

4µN = ρπ 

(6.120)

The equivalent damping ratio becomes ceq 2µN = ξeq = √ πkρβ 2 km

(6.121)

The displacement response is now given by Equation 6.27, where the amplitude ρ is obtained from Equation 6.28 with ξeq substituted for ξ ρ=

p0 1  k (1 − β 2 )2 + (4µN/πkρ)2

On solving Equation 6.122 for ρ, we get % p0 1 − (4µN/πp0 )2 ρ= k (1 − β 2 )2

(6.122)

(6.123)

Equation 6.123 is valid provided that the ratio of friction force to the amplitude of applied load, µN/p0 , is less than π/4. For a larger friction force, the numerator in Equation 6.123 becomes imaginary and the method breaks down. In fact, for reasonable accuracy µN/p0 must be less than about 12 .

Forced harmonic vibrations: Single-degree-of-freedom system

267

Figure 6.29 (a)Variation of amplitude ratio with frequency ratio and Coulomb friction; (b) variation of phase angle with frequency ratio and Coulomb friction.

Amplitude ratio ρ/(p0 /k) is plotted in Figure 6.29a as a function of β for several different values of µN/p0 . It is apparent from the plots and from Equation 6.123 that, at resonance, friction damping does not limit the amplitude, which therefore becomes unbounded.

268

Dynamics of structures

Phase angle φ is obtained from Equation 6.28b with ξ = ξeq and ρ given by Equation 6.123 tan φ = ± 

4µN/πp0 1 − (4µN/πp0 )2

(6.124)

where the positive sign applies for β < 1 and the negative sign for β > 1. The phase angle is independent of the frequency ratio but changes sign as β passes through 1. The variation of phase angle with the frequency ratio and µN/p0 is shown in Figure 6.29b. 6.16

MEASUREMENT OF DAMPING

As stated earlier, the mass and stiffness of a dynamic system can be determined from its physical characteristics. Mass can be calculated from the known geometry and the mass density. Stiffness is determined if the geometry and material properties are known. However, it is difficult or impractical to relate damping resistance to known or measurable physical characteristics. An estimate of damping resistance can, however, be obtained by experimental measurement of the response of the structure to a given excitation. From a practical point of view, harmonic excitation is easiest to impart. It is for this reason that a discussion of the measurement of damping resistance is appropriately included in a section on response to harmonic excitation. A number of different experimental techniques are available. Some of these are described in the following paragraphs. Most techniques are based on the assumption that damping is of viscous type. When damping is not viscous in nature, an equivalent viscous damping is usually obtained.

6.16.1 Free vibration decay In this method, which is related to free vibration described in Chapter 5, the system is excited by any convenient means and then allowed to vibrate freely. Measurements of displacement amplitudes are then made on a few successive peaks. For free vibration with viscous damping, if ui denotes the displacement at any peak and un+i is the displacement n cycles later, then, as shown in Section 5.3.5, the following relationship holds ui e−ξ ωti = −ξ ω(2πn/ω +t ) i d un+i e = e2π nξ ω/ωd √ 2 = e2π nξ/ 1−ξ

(6.125)

Viscous damping ratio ξ is obtained from Equation 6.125 ξ=

δn 4π 2 n2

+ δn2

(6.126)

where δn = ln(ui /un+i ). For light damping, Equation 6.126 reduces to ξ = δn /2π n.

Forced harmonic vibrations: Single-degree-of-freedom system

269

In the determination of damping resistance through an observation of the free vibration decay, the only measurements required are those of relative displacements, and the instrumentation required is therefore fairly simple. When damping is hysteretic, the damping force is defined as in Equation 5.56. For free vibration, the vibration frequency is approximately equal to ω, provided that damping is light. The equivalent damping constant ch and damping ratio ξh are therefore given by ηk ω ch η ξh = √ = 2 2 km

ch =

(6.127)

The hysteretic damping coefficient η is obtained by using Equation 6.126 with ξ replaced by ξh , so that η=  ≈

2δn 4π 2 n2

+ δn2

δn πn

(6.128)

6.16.2 Forced-vibration response Damping resistance can also be obtained by measuring the displacement response of the system to a harmonic excitation. The system is excited by a series of harmonic forces with closely spaced frequencies spanning the resonant frequency. The displacement response to each excitation is measured so that the entire response curve in the vicinity of resonance is obtained. The properties of the response curve can be used in a number of different ways to obtain an estimate of the damping. Resonant response At resonance, that is, when β = 1, the phase angle φ = π/2. Resonance can therefore be detected by measuring the phase angle and progressively adjusting the exciting frequency until φ is 90◦ . If the measured displacement amplitude at resonance is denoted by ur , then ur 1 = p0 /k 2ξ

(6.129a)

so that ξ=

1 p0 2ur k

(6.129b)

The application of Equation 6.129 requires that the static displacement produced by a force of the same amplitude as the dynamic force be ascertained. This can be done in a separate test.

270

Dynamics of structures

Measurement of the phase angle may be somewhat difficult. Therefore, as an alternative the resonance curve is obtained in the vicinity of resonance and the peak response up is measured. For viscous damping up 1 =  p0 /k 2ξ 1 − ξ 2

(6.130)

Equation 6.130 can be used to obtain ξ from the measured value of up and the static deflection produced by a force of amplitude p0 . For light damping ξ 2 is negligible in comparison to 1 and ξ = (p0 /k)(1/2up ). It should be noted that if the damping is hysteretic in nature, up and ur are identical and are given by up = ur =

1 p0 η k

(6.131)

6.16.2.1 Width of response curve and half-power method. The width of response curve near resonance can be used to obtain an estimate of the damping. The method involves the measurement of frequencies at which the phase angle is ±π/4, or tan φ = ±1. Referring to Figure 6.30, one such frequency will be below the resonant frequency and the other above it. The two values of β are obtained from Equation 6.28b 2ξβ1 =1 1 − β12

(6.132a)

2ξβ2 = −1 1 − β22

(6.132b)

Equations 6.132a and b give 1 − β12 − 2ξβ1 = 0

(6.133a)

1−

(6.133b)

β22

+ 2ξβ2 = 0

Subtracting Equation 6.133a from Equation 6.133b, we obtain 1 1 ξ = (β2 − β1 ) = 2 2




2 − 1 ω

 (6.134)

Equation 6.134 can be used to obtain ξ provided that 2 and 1 have been measured and natural frequency ω is either known or can be estimated. The foregoing method relies on the ability to measure the phase angle, which may require sophisticated instrumentation. Another property of the response curve whose use does not rely on the measurement of phase angle is therefore sometimes employed in estimating the damping ratio. If the response curve in the vicinity of resonance has

Forced harmonic vibrations: Single-degree-of-freedom system

271

Figure 6.30 Use of response curve in the vicinity of resonance to determine the damping ratio.

√ been plotted, the frequencies at which the amplitude is 1/ 2 times that at the peak can be measured. As shown in Figure 6.30, there will be two such frequencies. Let these frequencies be denoted by A and B and the corresponding frequency ratios, by βA and βB . For a small amount of damping, the peak response is (p0 /k)(1/2ξ ). Hence βA and βB are obtained from the equation 1 1 p0 p0 1 =  √ 2 2ξ k k 2 (1 − β )2 + (2ξβ)2

(6.135)

On solving Equation 6.135 for β, we get  βA2 = (1 − 2ξ 2 ) − 2ξ 1 + ξ 2  βB2 = (1 − 2ξ 2 ) + 2ξ 1 + ξ 2

(6.136)

For small values of ξ , Equation 6.136 can be simplified as follows βA2 ≈ 1 − 2ξ − 2ξ 2 βB2 ≈ 1 + 2ξ − 2ξ 2

(6.137a)

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Dynamics of structures

or βA ≈ 1 − ξ − ξ 2 βB ≈ 1 + ξ − ξ 2

(6.137b)

Equation 6.137b gives ξ=

1 (βB − βA ) 2

(6.138)

Equation 6.138 is similar to Equation 6.134. In fact, for small amounts of damping, frequencies corresponding to half power points, βA and βB are very close to those corresponding to phase angles of ±π/4, that is, β1 and β2 . Energy loss per cycle Instead of using the response curve obtained for a range of frequencies, it is possible to determine the damping resistance by running a harmonic excitation test at one specified frequency provided that the complete force–displacement relationship can be measured. When damping is viscous, the applied force versus displacement relationship will be a skewed ellipse whose intercept on the ordinate is equal to cρ. Since  and ρ are both known, c can be determined by measuring the intercept. For damping that is not viscous in nature, the force–displacement relationship will not in general be an ellipse. An equivalent damping constant can still be determined by measuring WD , the area enclosed by the force–displacement relationship, and equating it to the theoretical value of energy loss per cycle in viscous damping. Thus from Equation 6.99, WD = ceq πρ 2

(6.139a)

or ceq =

WD π ρ 2

(6.139b)

The damping ratio ξeq is given by ceq ξeq = √ 2 km

(6.140a)

or ξeq =

ceq ω 2k

(6.140b)

To use Equation 6.140a, m must be estimated from the geometry and mass density, and k can either be estimated or measured.

Forced harmonic vibrations: Single-degree-of-freedom system

273

Figure 6.31 Static force–displacement relationship.

Stiffness k can be determined by running a static load test. In such a test, if as shown in Figure 6.31, the area bounded by the force–displacement relationship, the displacement axis and an ordinate through amplitude ρ is denoted by Ws , we have Ws =

1 2 ρ k 2

(6.141)

Hence k=

2Ws ρ2

(6.142)

If it is desired that the damping ratio be obtained without having to estimate the mass, the test should be run at resonance. Resonance is achieved by adjusting the input frequency until the phase angle is 90◦ . The input frequency is now equal to the natural frequency and Equation 6.139b becomes ceq =

WD π ωρ 2

(6.143)

Substituting Equations 6.140b and 6.142 in Equation 6.143, we get ξeq =

WD 4π Ws

(6.144)

It is of interest to note that, at resonance, the viscous damping force exactly balances the applied force, and the force–displacement relationship is the ellipse shown in Figure 6.24c.

274

Dynamics of structures

Example 6.8 A water tower with a total weight of 50 kips is represented by a single-degree-of-freedom system. The total stiffness of the supporting columns is estimated at 52 kips/in. In a test the tower is subjected to a harmonic excitation at a frequency of 10 rad/s and the force displacement relationship is obtained at steady-state. The amplitude of displacement is measured as 2 in. and the energy loss per cycle as 65.2 kip · in. (a) (b) (c)

If the damping is considered to be viscous in nature, determine c and ξ . If the damping is considered to be hysteretic, determine η. If the test is rerun at 20 rad/s, and the amplitude of displacement is still 2 in., what would be the energy loss if the damping is truly viscous in nature? What would the energy loss be, if the damping were hysteretic?

Solution (a)

For viscous damping, the energy loss per cycle is given by Equation 6.96 WD = cπρ 2

(a)

With  = 10 rad/s, ρ = 2 in., and WD = 65.2 kip · in., Equation a gives c = 0.519

kip · s in.

The natural frequency of the system is obtained from ) ) k 52 × 386.4 ω= = m 50 = 20.05 rad/s The damping ratio is given by ξ=

(b)

0.519 × 386.4 c = 2 × 50 × 20.05 2mω = 0.1

In this case the energy loss is given by Equation 6.104 W = ηkπρ 2

(b)

Substituting for W, k, and ρ in Equation b we get η = 0.1 (c)

If the damping is viscous, the energy loss per cycle at a test frequency of 20 rad/s will be given by W = cπρ 2 = 0.519 × π × 20 × 4 = 130.4 kip · in. If damping were hysteretic, the energy loss per cycle will not change and will still be 65.2 kip · in.

Forced harmonic vibrations: Single-degree-of-freedom system

275

SELECTED READINGS Clough, R.W., and Penzien, J., Dynamics of Structures, 2nd Edition, McGraw-Hill, New York, 1993. Crede, C.E., Vibration and Shock Isolation, John Wiley, New York, 1951. Crede, C.E., Shock and Vibration Concepts in Engineering Design, Prentice Hall, Englewood Cliffs, 1965. Harris, C.M., (ed.) Shock and Vibration Handbook, 4th Edition, McGraw-Hill, New York, 1996. Eller, E.E., and Whittier, R.M., “Piezoelectric and Piezoresistive Transducers’’, Shock and Vibration Handbook, 4th Edition, (Ed. C.M. Harris), McGraw-Hill, New York, 1996. Hudson, D.E., “Ground Motion Measurements’’, in Earthquake Engineering (Ed. R. L. Weigel), Prentice-Hall, Englewood cliffs, N.J., 1970. Hudson, D.E., Reading and Interpreting Strong Motion Accelerograms, Earthquake Engineering Research Institute, Berkeley, Calif.,1979. Myklestad, N.O., “The Concept of Complex Damping’’, Journal of Applied Mechanics, Vol. 19, 1952, pp. 284–286. Westermo, B., and Udwadia, F., “Periodic Response of a Sliding Oscillator System to Harmonic Excitation’’, Earthquake Engineering and Structural Dynamics, Vol. 11, 1983, pp. 135–146.

PROBLEMS 6.1

Two uniform parallel beams AB 1.2 m long, weighing 20 kg each, are hinged at A and supported at B by a spring of stiffness 4.33 N/mm as shown in Figure P6.1. The beams carry a flywheel of mass 132 kg attached at D, 1.0 m away from the hinge. The flywheel is rotating at 200 RPM and its center of gravity is offset 3 mm from the axis of rotation. Find (a) the natural frequency of the system, (b) the maximum vertical movement of end B of the beams. Neglect damping.

Figure P6.1 6.2

A generator is mounted on a bed plate supported on four spring-dampers, the springs each having stiffness 20, 000 lb/in. The damping present is estimated to be 20% of critical and the total weight supported is 2240 lb. Measurements show a maximum amplitude of vertical motion of 0.0025 in. at a speed of 2000 RPM of the generator. Calculate the force transmitted to the foundation, assuming the motion to be simple harmonic.

6.3

An electric motor weighing 1160 N is attached to a floor beam which deflects 0.76 mm under the weight. The armature of the motor weighs 40 kg. As the motor is run up gradually to operating speed of 1600 RPM, it is observed that the maximum amplitude of oscillation is 2.5 mm, decreasing to 0.5 mm at the operating speed. Calculate the damping coefficient and the eccentricity of the armature from its axis of rotation.

276

Dynamics of structures

6.4

A vibration test machine consists of a 1200-lb payload mounted on a beam arrangement. The static deflection is noted to be 0.386 in. Damping is small, but the oscillation of the system was observed to decrease from 0.2 in. to 0.05 in. in 18 cycles. If the supports of the beam can be given a harmonic displacement of amplitude 0.1 in. at the natural frequency of the system, estimate the expected amplitude attained by the payload.

6.5

A machine weighing 900 N is supported on springs of total stiffness 700 N/mm. Unbalance results in a disturbing force of 360 N at a speed of 3000 RPM. Damping is estimated at 20% of critical value. Determine (a) the amplitude of motion (b) the transmissibility (c) the transmitted force

6.6

An electronic instrument of mass 45 kg is to be attached to an aircraft bulkhead. Vibration measurements on the bulkhead reveal a motion consisting of three sinusoidals: (i) frequency 20 Hz and amplitude 0.018 mm, (ii) frequency 35 Hz and amplitude 0.030 mm, (iii) frequency 53 Hz and amplitude 0.045 mm. The installation specifications call for a suspension system that will limit the amplitude of vibration to not more than 0.004 mm. What spring rate would you recommend for the suspension?

6.7

A seismic instrument designed to measure vertical acceleration consists of a light beam AB pivoted at A through a torsional spring of constant 20 Nm (Fig. P6.7). A small body of mass 0.6 kg is fixed to the beam at end B and a viscous damper is attached at C. The pointer BED, pivoted to the 0.6 kg mass at B and to the instrument casing at E, moves over the scale at D as shown in the figure. The viscous damper is adjusted to provide a damping of 0.7 times the critical. Obtain the undamped natural frequency of the instrument and the damping coefficient of the viscous damper. The instrument is calibrated to read the support acceleration correctly at an input frequency of 1 Hz. Obtain the calibration factor. If the support acceleration is 3 m/s2 at a frequency of 5 Hz what will be the instrument reading and what is the percentage error?

Figure P6.7 6.8 6.9

For the seismic instrument of Problem 6.7 what are the phase differences between the measured and input signals for each of the two input frequencies: 1 Hz and 5 Hz? An accelerometer has a natural frequency of 40 rad/s and a damping ratio of 0.6. It is calibrated to read correctly the input acceleration at very low frequencies. Find the calibration factor. An input signal given by u¨ g = 4 sin(4πt) + 3 sin(5πt) m/s2 is supplied to the accelerometer. Plot the output signal produced by the instrument. What are the measured amplitudes of the two components of the given signal.

Forced harmonic vibrations: Single-degree-of-freedom system 6.10

6.11

277

A displacement meter has a natural frequency of 3 Hz and damping ratio of 0.7. It is calibrated to read correctly the displacements at very high frequencies. What are the errors in readings at input frequencies of 4.5 Hz, 6 Hz, and 9 Hz? What is the phase shift in each case? A rotating unbalanced mass m0 has an eccentricity e and frequency  rad/s. It is supported by a system of springs such that the natural frequency of vibration is ω. Show that with hysteretic damping, the amplitude of steady state response is given by ρ=

em0 2 β m 1

{(1 − β 2 )2 + η2 } 2

where β = /ω is the frequency ratio, and η is the  hysteretic damping ratio. Hence prove that the maximum amplitude occurs when β = 1 + η2 and is given by 1 ρmax 1 + η2 = em0 /m η 6.12

To obtain an estimate of the damping in the system shown in Problem 6.1, end B is displaced 32 mm and the system is allowed to vibrate freely. The amplitude decays to 6 mm in 5 cycles. If the damping is assumed to be viscous in nature, what will be the maximum amplitude of B due to the rotation of eccentric flywheel.

6.13

Solve Problem 6.12 assuming the damping to be hysteretic.

6.14

Using an eccentric mass shaker, a dynamic test was run on a structure in which the exciting frequency was varied over a range of values. For each frequency, the response amplitude as well as the phase angle were measured and the following data was obtained. The estimated natural frequency of the structure is 2 Hz. Plot a frequency versus amplitude curve; then using the half power method, obtain an estimate of the damping. It will be readily seen that the graphical relationship obtained by taking the amplitude ρ and the phase angle φ as polar coordinates is a Nyquist plot. Draw such a plot from the data supplied. By estimating the values of 1 and 2 corresponding to φ = 45◦ and 135◦ from the Nyquist plot, obtain an estimate of the damping. Of the two estimates of damping obtained by you, which do you think is more reliable.  (Hz) 0.0 0.4 0.8 1.2 1.6 1.8 2.0 2.2 2.4 2.8 3.2 3.6 4.0

6.15

ρ (mm) 5.00 5.20 5.60 7.70 12.70 19.10 25.00 16.40 10.00 5.00 3.10 2.20 1.70

φ (deg) 0.0 2.4 5.4 10.6 24.0 43.5 90.0 133.2 151.4 163.7 168.4 170.9 172.4

Show that the Nyquist plot for the displacement of a damped system with hysteretic 1 1 damping is a circle with center (0, 2η ) and radius 2η . Also show that half power points

278

Dynamics of structures occur exactly at phase angles 45o and 135o . Obtain a Nyquist plot from the following data and hence calculate the damping factor η β 0.0 0.2 0.4 0.6 0.8 0.9 1.0 1.1 1.2 1.4 1.6 1.8 2.0

6.16

ρ (p0 /k)

0.98 1.02 1.16 1.49 2.43 3.63 5.00 3.45 2.07 1.02 0.64 0.44 0.33

φ (deg) 11.3 11.8 13.4 17.4 29.1 46.5 90.0 136.4 155.6 168.2 172.7 174.9 176.2

In order to measure the characteristics of friction dampers used in earthquake-resistant construction, laboratory tests are conducted on a scaled model of a single-story frame as shown in Figure P6.16. In a free vibration test, the amplitude of motion of the frame reduces from 3/4 in. to 1/64 in. in 6 cycles of motion which take 1.5 s. The lateral stiffness of the frame is estimated at 20 kips/in. An eccentric mass shaker mounted on the frame is then run at a frequency of 6 Hz. If the unbalanced force is 3 kips, find the maximum amplitude of vibration.

Figure P6.16

6.17

The lateral displacement of the frame shown in Figure P6.17 are measured under a quasistatic cyclic load varying from −12 to 12 kips. The experimental results are shown below. Plot the hysteretic loop of the frame. Obtain η by estimating the area enclosed by the loop. If a harmonic force of amplitude 12 kips is applied to the frame at resonant frequency, what would be the maximum displacement of the frame?

Forced harmonic vibrations: Single-degree-of-freedom system

Figure P6.17

Lateral force fSt (kips) 3.18 5.13 7.04 8.89 10.66 12.00 9.34 7.11 4.96 2.87 0.82 −1.20 −3.18 −5.13 −7.04 −8.89 −10.66 −12.0 −9.34 −7.11 −4.96 −2.87 −0.82 1.20

Displacement u (in.) 0.2 0.4 0.6 0.8 1.0 1.2 1.0 0.8 0.6 0.4 0.2 0.0 −0.2 −0.4 −0.6 −0.8 −1.0 −1.2 −1.0 −0.8 −0.6 −0.4 −0.2 0.0

279

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Chapter 7

Response to general dynamic loading and transient response

7.1

I NTRODUCTION

The analysis of response to a forcing function that is neither harmonic nor periodic is comparatively more complex. However, for linear systems a general method exists which is conceptually quite straightforward. This method is based on the ability to obtain mathematical expressions for the response of the system to an impulsive force. A general forcing function is expressed as the sum of a series of impulses and the response to the applied force is obtained by superposing the responses to the individual impulses. Superposition, in fact, involves the evaluation of an integral, known as a convolution integral or Duhamel’s integral, either mathematically or by numerical means of integration. An important class of nonperiodic loads are those that act for a relatively short duration of time. Such loads are also known as impulsive loads or shock loads, and may have considerable significance in the design of certain systems. Examples are: loads generated due to a blast or explosion, and dynamic loads in automobiles, traveling cranes, and other mobile machinery. The response produced by such loads is transient in nature and decays rapidly after the load has ceased to act. However, from the point of view of engineering design, the duration of transient response is of no particular importance; rather, it is the maximum displacement and stress attained during the response that is of interest. Because of the short duration of response, damping does not have a significant influence and can reasonably be ignored in the analysis.

7.2

RESPONSE TO AN IMPULSIVE FORCE

An impulsive force is a large force that acts for a very short duration of time. Although the magnitude of such a force may be infinitely large, its time integral, that is, the area enclosed by the force–time curve and the time axis, is finite. The time integral is referred to as the impulse of the force. Referring to Figure 7.1, the magnitude of impulse, denoted by I, is obtained from  I=

t+ε

p(t) dt

(7.1)

t

where ε is the very small interval of time during which the impulsive force is acting.

282

Dynamics of structures

Figure 7.1 Impulse of an impulsive force.

Mathematically, an impulsive force can be expressed in terms of the delta function. Function δ(t) represents a delta function centered at time t = 0. The magnitude of the function is zero at all times other than zero. At t = 0, the function value is infinitely large. However, the area enclosed by the delta function and the time axis is finite and is equal to unity. Thus an impulsive force applied at time t = 0 and having an impulse equal to I can be represented by I δ(t). Analogously, function I δ(t − τ ) represents a force of impulse I acting at time t = τ . Newton’s second law of motion states that the action of an impulsive force on a mass results in a change in the velocity of the mass and hence in its momentum, the change in momentum being equal to the impulse of the impulsive force. Thus, representing the change in velocity by v, mv = I

(7.2)

If the mass was initially at rest, it will have a velocity I/m after the action of the impulse. Suppose that a single-degree-of-freedom system at rest having a mass m, a restraining spring stiffness k, and negligible damping is subjected to an impulsive force with impulse I. The action of the impulsive force will set the system vibrating. The ensuing free-vibration response can be obtained by recognizing that the initial displacement is zero and the initial velocity is given by v0 = I/m. The resulting response is given by Equation 5.9a with u0 = 0: u=

I sin ωt mω

(7.3)

The response to an impulse of unit magnitude is called unit impulse response. It is denoted by h(t) and is given by h(t) =

1 sin ωt mω

(7.4)

Response to general dynamic loading and transient response

283

Figure 7.2 Representation of a general dynamic load by a series of impulses.

When damping is present, the unit impulse response is obtained from Equation 5.37 with u0 = 0 and v0 = 1/m h(t) =

7.3

1 −ωξ t e sin ωd t mωd

(7.5)

RESPONSE TO GENERAL DYNAMIC LOADING

The unit impulse response functions given by Equations 7.4 and 7.5 can be used to obtain the response to any general dynamic loading. The system being excited is assumed to be linear, so that the principle of superposition holds. The applied force can be viewed as the sum of a series of impulses such as the one shown shaded in Figure 7.2. The shaded impulse has a magnitude of p(τ ) dτ and is imparted at time t = τ . At any subsequent time t, when the elapsed time since the impulse is t − τ , the incremental response, du, due to the impulse under consideration is obtained from Equation 7.4 with t − τ replacing t: du =

p(τ ) dτ sin ω(t − τ ) mω

(7.6)

in which the system has been assumed to be undamped. The total response at time t is obtained by superposing the effects of all impulses from time τ = 0 to τ = t, giving u(t) =

1 mω



t

p(τ ) sin ω(t − τ ) dτ

(7.7)

0

In a similar manner, the response of a damped system is given by u(t) =

1 mωd



t 0

p(τ ) e−ξ ω(t−τ ) sin ωd (t − τ ) dτ

(7.8)

284

Dynamics of structures

Superposition integral equations 7.7 and 7.8 can be expressed in the general form  u(t) =

t

p(τ ) h(t − τ ) dτ

(7.9)

0

where an appropriate expression is used for the unit impulse function h. Equation 7.9 is known as the convolution integral or Duhamel’s integral. It provides a general method for the analysis of the response of a linear system to an arbitrary loading. It also forms the basis for the development of the Fourier transform method of analysis described in Chapter 9. When p(τ ) is a simple mathematical function, closed-form evaluation of the integral in Equation 7.9 is possible; in other cases, a numerical technique must be used in the evaluation. Some of the numerical techniques are discussed in Chapter 8. It should be noted that in obtaining Equation 7.9, the system was assumed to be at rest at time t = 0. If the system starts with initial condition u0 and v0 , the freevibration component of the response given by Equation 5.9a (undamped system) or Equation 5.37 (damped system) must be added to Equation 7.9 to obtain the total response. Thus for a damped system
 v0 + ωξ u0 −ωξ t u(t) = e u0 cos ωd t + sin ωd t ωd (7.10)  t 1 −ξ ω(t−τ ) + p(τ ) e sin ωd (t − τ ) dτ mωd 0 In the following sections, we present several examples of the use of Duhamel’s integral in obtaining the response to a general dynamic load.

7.4

RESPONSE TO A STEP FUNCTION LOAD

A step function load is a suddenly applied load that remains constant after application. It is a simple example of a nonperiodic load. Figure 7.3 shows a step function load applied at time t = 0. The equation of motion of a single-degree-of-freedom damped system subjected to a step load of magnitude P0 is given by mu¨ + cu˙ + ku = P0

(7.11)

Figure 7.3 Step function load.

Response to general dynamic loading and transient response

285

The complementary solution of Equation 7.11 is given by Equation 5.36. For a particular solution we use the method of trials and assume for our trial solution u = G. Substitution in Equation 7.11 shows that G = P0 /k. The complete solution then becomes u = e−ωξ t (A cos ωd t + B sin ωd t) +

P0 k

(7.12)

in which A and B are arbitrary constants to be determined from initial conditions. When the system starts from rest, u0 = v0 = 0 and the arbitrary constants A and B are easily shown to be A=−

P0 k

(7.13)

P0 ωξ B=− k ωd The resulting solution becomes u=


  ωξ P0 sin ωd t 1 − e−ωξ t cos ωd t + k ωd

(7.14)

The dynamic load factor u/(P0 /k), which gives the ratio of the dynamic displacement to the static displacement under load P0 , is plotted in Figure 7.4 as a function of t/T, where T = 2π/ω is the natural period of the system, for several different values of the

Figure 7.4 Response to a step function load.

286

Dynamics of structures

damping ratio ξ . The dynamic load factor attains its maximum value of 2 when the damping is negligible. For finite damping it is always less than 2. The time at which the maximum occurs can be obtained by equating the differential of Equation 7.14 to zero. Thus, if tp represents the time at peak value,   P0 −ωξ tp (ωξ )2 + ωd sin ωd tp = 0 e k ωd

(7.15)

or tp =

nπ ωd

n = 0, 1, 2, . . .

(7.16)

For n = 0, tp = 0, and substitution into Equation 7.14 gives u = 0, which represents a minimum. The first peak occurs when n = 1 and tp = π/ωd . Other values of n represent subsequent peaks in the displacement curve where the magnitude of displacement is less than that at the first peak. The maximum is thus obtained by substituting t = π/ωd in Equation 7.14. P0 (1 + e−ωξ π/ωd ) k √ P0 2 = (1 + e−πξ/ 1−ξ ) k

umax =

(7.17)

The peak response ratio is thus a function of only the damping ratio. Figure 7.5 shows the relationship between umax /(P0 /k) and ξ . As expected, the maximum value of the response ratio is 2 for ξ = 0.

Figure 7.5 Maximum response to a step function load.

Response to general dynamic loading and transient response

287

The response to step function can also be obtained by using the Duhamel’s integral. u(t) =

1 mωd



t

P0 e−ωξ (t−τ ) sin ωd (t − τ ) dτ

(7.18)

0

Evaluation of the integral in Equation 7.18 will provide an expression for the response that is identical to Equation 7.14.

7.5

RESPONSE TO A RAMP FUNCTION LOAD

A ramp function load is a load that increases linearly with time as shown in Figure 7.6a. Mathematically, it can be expressed as p(t) =

P0 t t1

(7.19)

The response of a single-degree-of-freedom system to a ramp function load is obtained by substituting for p(τ ) in the Duhamel’s integral, 1 u(t) = mwd



t 0

P0 τ −ωξ (t−τ ) e sin ωd (t − τ ) dτ t1

(7.20)

For an undamped system, the Duhamel’s integral simplifies to  t P0 τ 1 sin ω(t − τ ) dτ mω 0 t1 
sin ωt P0 t − = k t1 ωt1

u(t) =

Figure 7.6 Response to a ramp function load: (a) ramp function load; (b) response.

(7.21a) (7.21b)

288

Dynamics of structures

Equation 7.21b has been plotted in Figure 7.6b for a specific value of t1 /T. The response is seen to oscillate about the average value P0 t/(kt1 ). The velocity response of the system to the ramp function load is obtained by taking the differential of Equation 7.21. u(t) ˙ =

P0 (1 − cos ωt) kt1

(7.22)

Since cos ωt ranges between −1 and +1, the velocity is either zero or positive. This implies that the system displacement is either stationary or is increasing with time. At some stage the displacement reaches the elastic limit of the restraining spring; the solutions given by Equations 7.20 through 7.22 are then no longer valid. 7.6

RESPONSE TO A STEP FUNCTION LOAD WITH RISE TIME

A load that rises linearly to P0 in time t1 and then remains constant at P0 is shown in Figure 7.7a. Such a load is commonly known as a step function load with a rise time. It can be viewed as the superposition of a ramp function load applied at time t = 0, and an equal but negative ramp function load applied at time t = t1 .

Figure 7.7 (a) Step function load with rise time; (b) response to step function load with rise time.

Response to general dynamic loading and transient response

289

The response of an undamped system to a ramp function load applied at t = 0 is given by Equation 7.21b. The response to a ramp function applied at time t = t1 is P0 u(t) = k



t − t1 sin ω(t − t1 ) − t1 ωt1

 t > t1

(7.23)

The response to the step function with rise time can be obtained from Equations 7.21 and 7.23. For t ≤ t1 , Equation 7.21b applies, for t > t1 , the response is obtained by subtracting Equation 7.23 from Equation 7.21b.  . /  P0 t − sin ωt ωt1 k . t1 u(t) = P  0 1 − sin ωt + ωt1 k

sin ω(t−t1 ) ωt1

/

t ≤ t1

(7.24a)

t > t1

(7.24b)

The history of response is shown in Figure 7.7b for a specific value of ωt1 . In the second era of response, that is, for t > t1 , the system executes a simple harmonic motion about the position of equilibrium. For t < t1 the response is similar to that for a ramp function load, so that the displacement continues to grow with time and the velocity at t = t1 is either zero or positive. This implies that the maximum value of response occurs either at t = t1 or in the era t > t1 . The time, tp , at which the maximum occurs is obtained by equating the time differential of Equation 7.24b to zero. Thus P0 {−cos ωt + cos ω(t − t1 )} = 0 kt1

(7.25)

After simplification, Equation 7.25 gives tan ωt = tan

ωt1 2

(7.26a)

or tp =

nπ t1 + ω 2

n = 0, 1, 2, . . .

(7.26b)

Obviously, the value of n in Equation 7.26b should be chosen so that tp ≥ t1 : Substitution of t = tp in Equation 7.24b gives the following value for the maximum response:

umax

  2 sin(nπ − ωt1 /2) P0 1+ = k ωt1

(7.27)

290

Dynamics of structures

Figure 7.8 Response spectrum for a step function load with rise time.

The true maximum will be obtained by selecting a value of n such that the second term within the braces in Equation 7.27 is positive, so that umax =

  P0 2|sin(ωt1 /2)| 1+ k ωt1

(7.28)

It is of interest to note that for the special case when ωt1 = 2nπ the displacement at t = t1 is P0 /k as shown by Equation 7.21b, while the velocity is zero as shown by Equation 7.22. The displacement in the era t > t1 , given by Equation 7.24b, remains constant at P0 /k and the system does not oscillate. The response ratio umax /(P0 /k) has been plotted as a function of ωt1 /2π = t1 /T in Figure 7.8. Plots of the type of Figure 7.8 which show the relationship between the maximum value of a response parameter and a characteristic of the system are called response spectra. The system characteristic used in a response spectrum is either the natural frequency or the natural period. In Figure 7.8 the period T, or rather its inverse, may be considered as the variable system characteristic, while the response parameter is the maximum displacement. Availability of response spectra often greatly simplifies the design of a system for a given loading. It is clear from both Equation 7.27 and Figure 7.8 that as t1 /T increases, the maximum response tends to a value equal to the static displacement under load P0 . In other words, if the load is applied very gradually, the response is essentially static. Example 7.1 A suddenly applied load P0 that decays exponentially after application is shown in Figure E7.1a. Obtain the response of an undamped system to such a load.

Response to general dynamic loading and transient response

291

Figure E7.1 (a) Exponentially decaying load; (b) response to exponentially decaying load.

Solution The equation of motion is mu¨ + ku = P0 e−at

(a)

and the response is obtained by using the Duhamel’s integral, P0 u(t) = mω



t

e−aτ sin ω(t − τ ) dτ

(b)

0

in which it has been assumed that the system starts from rest. Evaluation of integral in Equation b gives u=

a  P0 sin ωt − cos ωt + e−at 2 2 k(1 + a /ω ) ω

(c)

The displacement response has been plotted in Figure E7.1b. It shows that the system oscillates with decreasing amplitude. As t increases, the term e−at becomes very small and the system attains a steady-state harmonic motion whose amplitude is given by ρ=

P0 1  k 1 + a2 /ω2

Example 7.2 A blast-induced pressure wave striking a single-story building is represented by p = 100(e−10t − e−100t ), where p is the pressure in psf. The building face subject to the wind pressure has an area of 144 ft2 . The mass of the building, assumed concentrated at the floor level, is 0.1 kip · s2 /in. and the total stiffness of the columns supporting the floor is 350 kips/in. Find the displacement response of the building and the maximum base shear induced by the pressure wave.

292

Dynamics of structures

Figure E7.2 (a) Blast-induced pressure wave; (b) response of single-story building to blast-induced pressure wave.

Solution As seen from Figure E7.2a, the superposition of two exponentially decaying waves can be used effectively to represent a blast-generated wave. The force applied at the floor level of the building due to the pressure wave is given by P=

1 × 144 × 100(e−10t − e100t ) × 10−3 kips 2

(a)

The total response is obtained by summing the response to the two component waves. The response to each component wave is obtained by an expression of the form of Equation c of

Response to general dynamic loading and transient response

293

Example 7.1, so that u=

a  P0 −at sin ωt − cos ωt + e k(1 + a2 /ω2 ) ω 
b P0 −bt sin ωt − cos ωt + e − k(1 + b2 /ω2 ) ω

(b)

In this case, k = 350 kips/in. ) 350 = 59.16 rad/s ω= 0.1 P0 = 7.2 kips a = 10 b = 100

On substituting the values, we get u=




 10 sin ωt − cos ωt + e−10t 59.16
 100 7.2 sin ωt − cos ωt + e−100t − 350(1 + 2.857) 59.16 7.2 350(1 + 0.028)

(c)

or u = −0.0056 sin ωt − 0.01467 cos ωt + 0.02e−10t − 0.0053e−100t

(d)

The response given by Equation d has been plotted in Figure E7.2b. The maximum displacement is 0.02685 in. and hence the maximum base shear = kumax is 9.4 kips.

7.7

RESPONSE TO SHOCK LOADING

In the preceding sections, examples were presented of response to nonperiodic loading. We now discuss several cases of response to shock loading or impulsive loading. As the name implies, the loading in these cases is of short duration. As a result, the maximum value of the response is not significantly affected by the presence of damping. The effect of damping is therefore ignored in the presentation.

7.7.1 Rectangular pulse A rectangular pulse load of duration t1 seconds is shown in Figure 7.9. For t ≤ t1 , the response of the system is the same as that to a step function load. Thus setting ξ = 0 in Equation 7.14, we get u(t) =

P0 (1 − cos ωt) t ≤ t1 k

(7.29)

294

Dynamics of structures

Figure 7.9 Rectangular pulse load.

At time t = t1 the displacement and velocity of the system are given by P0 (1 − cos ωt1 ) k P0 ω u(t ˙ 1) = (sin ωt1 ) k

u(t1 ) =

(7.30a) (7.30b)

In the second era of response, that is, for t > t1 , the system is undergoing freevibration with initial conditions given by Equations 7.30. The response in this era is therefore obtained from Equation 5.9a. u = u(t1 )cos ωt¯ +

u(t ˙ 1) sin ωt¯ ω

(7.31)

where t¯ = t − t1 is the time measured from the beginning of the free-vibration era. Substitution of appropriate values in Equation 7.31 gives u=

P0 {cos ω(t − t1 ) − cos ωt} k

t > t1

(7.32)

As an alternative, the rectangular pulse can be obtained by a superposition of a step function of magnitude P0 applied at t = 0 and another step function of magnitude −P0 applied at time t = t1 . The response to the second step function is u=−

P0 {1 − cos ω(t − t1 )} k

(7.33)

The response in the second era is then obtained by taking the sum of Equations 7.29 and 7.33, which leads to Equation 7.32. The histories of displacement response obtained from Equations 7.29 and 7.32 for four different values of the ratio t1 /T are shown in Figure 7.10. The static displacement produced by the rectangular pulse load is also shown in that figure by dashed lines. For t1 /T = 0.25 the maximum response occurs in the free-vibration era, and the maximum

Response to general dynamic loading and transient response

295

Figure 7.10 Time histories of response to a rectangular pulse load.

value is 1.414 P0 /k. For t1 /T = 0.5 the maximum response occurs at t = t1 , that is, at the end of the pulse, and the maximum value is 2 P0 /k. For the special case when ωt1 = 2nπ or t1 /T = n the displacement and velocity are both zero at t = t1 , and the motion ceases at this point. For both t1 /T = 1 and 1.5 the maximum response is 2P0 /k and occurs within the duration of the pulse. As seen from the response histories, the maximum value of response is attained in either the first or second vibration era, depending on the value of t1 . First, assuming that the maximum is achieved for t < t1 , the time at maximum, tp , is obtained by equating the differential of Equation 7.29 to zero. P0 ω sin ωtp = 0 k ωtp = nπ

n = 1, 3, 5, . . .

(7.34)

In Equation 7.34 n has been taken as being odd, since for even n the displacement is a minimum as will be evident from Equation 7.29. The maximum value of response is reached in the forced-vibration era provided that the smallest value of tp , obtained by taking n = 1, is less than t1 , that is tp ≡

π < t1 ω

(7.35a)

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Dynamics of structures

or t1 1 > T 2

(7.35b)

The corresponding value of the maximum response obtained from Equation 7.29 with t = tp is given by umax =

2P0 k

(7.36)

If the condition given by Equation 7.35 does not hold, the maximum will occur in the free-vibration era, and the time to maximum will be given by equating the differential of Equation 7.32 to zero. P0 ω{−sin ω(tp − t1 ) + sin ωtp } = 0 k

(7.37)

After some manipulation, Equation 7.37 gives tan ωtp = −cot

ωt1 2

(7.38a)

or tp = (2n + 1)

t1 π + 2ω 2

T t1 = (2n + 1) + 4 2

n = 0, 1, 2, . . .

(7.38b)

The maximum response is obtained by substituting the value of tp from Equation 7.38b for t in Equation 7.32, giving P0 ωt1 sin k 2 P0 π t1 = 2 sin k T

umax = 2

(7.39)

The response ratio umax /(P0 /k) has been plotted as a function of t1 /T in Figure 7.11. Figure 7.11 is another example of a response spectrum. When the loading is impulsive, as in this case, the response spectrum is usually referred to as a shock spectrum.

Response to general dynamic loading and transient response

297

Figure 7.11 Shock spectrum for a rectangular pulse load.

Figure 7.12 Triangular pulse load.

7.7.2 Triangular pulse Figure 7.12 shows a triangular pulse load of total duration t1 . The pulse can be represented as a superposition of three ramp functions indicated in the figure and given by p1 =

2P0 t t1

p2 = − p3 =

4P0 (t − t1 /2) t1

2P0 (t − t1 ) t1

t>0 t>

t1 2

(7.40)

t > t1

The first ramp function applied at time t = 0 has a slope of 2P0 /t1 , the second ramp function applied at time t = t1 /2 has a slope of −4P0 /t1 , and the third ramp

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Dynamics of structures

function applied at t = t1 has a slope of 2P0 /t1 . The response to ramp function p1 is directly obtained from Equation 7.21b. 2P0 u1 (t) = k




sin ωt t − t1 ωt1

 (7.41)

The response to function p2 is obtained from Equation 7.21b by substituting t − 12 t1 for t and multiplying the amplitude by −4. 4P0 u2 (t) = − k

!

sin ω(t − 12 t1 ) t − 12 t1 − t1 ωt1

" (7.42)

In a similar manner the response to function p3 is obtained from Equation 7.21b by substituting t − t1 for t and multiplying the amplitude by 2. u3 (t) =

2P0 k




sin ω(t − t1 ) t − t1 − t1 ωt1

 (7.43)

The response to the triangular pulse in the era 12 t1 < t ≤ t1 is obtained by superposing u1 (t) and u2 (t), while the response in the era t > t1 is obtained by superposing u1 , u2 , and u3 . Thus, we get    2P0 t ωt  − sin  t ωt k  1 1   . / 2 sin ω(t−t1 /2) ωt u(t) = 2Pk 0 1 − tt1 − sin + ωt ωt 1 1    2P . sin ωt 2 sin ω(t−t /2) sin ω(t−t ) /  1  0 − + − ωt1 1 ωt1 ωt1 k

t≤ t1 2

t1 2

< t ≤ t1

t1 < t

(7.44a) (7.44b) (7.44c)

The histories of displacement response obtained from Equations 7.44a, 7.44b, and 7.44c for six different values of the ratio t1 /T are shown in Figure 7.13. The static displacement produced by the triangular pulse load is also shown in that figure by dashed lines. For t1 /T = 0.25 the maximum response occurs in the free-vibration era, and the maximum value is 0.746 P0 /k. For t1 /T = 0.5 the maximum response occurs at t = t1 , that is, at the end of the pulse, and the maximum value is 1.273 P0 /k. For t1 /T = 1, 1.5, 2 and 3 the maximum occurs during the forced-vibration era. For the special case when t1 /T = 2n or ωt1 = 4nπ the displacement and velocity are both zero at t = t1 ; the motion ceases at this point and the maximum response equal to P0 /k occurs within the duration of the pulse at t = t1 /2. To obtain the shock spectrum, we first assume that the maximum response occurs in the free-vibration era t > t1 . Then the time to maximum, tp , is obtained by equating the time differential of Equation 7.44c to zero.  
 2P0 t1 − ω cos ω(tp − t1 ) = 0 −ω cos ωtp + 2ω cos ω tp − kωt1 2

(7.45)

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299

Figure 7.13 Time histories of response to a triangular pulse load.

After some manipulation, Equation 7.45 gives 
ωt1 =0 cos ωtp − 2

(7.46a)

or π tp = (2n + 1) 2ω +

= (2n + 1) T4 +

t1 2

t1 2

n = 0, 1, 2, . . .

(7.46b)

The value of the maximum response is obtained by substituting tp from Equation 7.46b for t in Equation 7.44c, which gives 
ωt1 4 umax (7.47) 1 − cos =± P0 /k ωt1 2 where the negative sign in Equation 7.44 applies for odd values of n. The maximum response will occur in the free-vibration era provided that tp obtained from Equation 7.46b is greater than t1 . Suppose now that tp obtained from Equation 7.46 with n = 0 is less than t1 , but that obtained with n = 1 is greater than t1 . The first peak in the free-vibration era is then negative. This implies that the system velocity at the end of the forced-vibration era is negative, and that a peak positive displacement was

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Dynamics of structures

Figure 7.14 Shock spectrum for a triangular pulse load.

reached during the forced-vibration era. Obviously, the peak in the forced-vibration era is going to be larger than any peak in the free-vibration era. This reasoning leads to the conclusion that if the maximum displacement is to occur in the free-vibration era, the tp obtained with n = 0 should be larger than t1 , so that the first peak in the free-vibration era is positive. Thus

tp =

T t1 + ≥ t1 4 2

or 1 t1 ≤ T 2

(7.48)

For t1 /T > 1/2, the maximum occurs in the forced-vibration era. From our discussion of the response to a ramp function load we know that during the period t ≤ t1 /2 when the force is rising the displacement continues to grow with time and the velocity at t = t1 /2 is either zero or positive. As a consequence, if the maximum occurs in the forced-vibration era it must occur in the period t1 /2 ≤ t ≤ t1 . The time to maximum is obtained by equating the differential of Equation 7.44b to zero. Substitution of this value of time in Equation 7.44b then gives the maximum response. The shock spectrum for the triangular pulse load is shown in Figure 7.14, in which the maximum response ratio umax /(P0 /k) has been plotted as a function of t1 /T.

Response to general dynamic loading and transient response

301

Figure 7.15 Sinusoidal pulse load.

7.7.3 Sinusoidal pulse The sinusoidal pulse shown in Figure 7.15 can be considered to be the superposition of two sine waves, the second of which begins at t = t1 seconds. The period Tf of each sine wave is 2t1 seconds, corresponding to a frequency of  = π/t1 rad/s. For t ≤ t1 , the response is obtained from Equation 6.8 u(t) =

P0 1 ( sin t − β sin ωt) t ≤ t1 k 1 − β2

(7.49)

where it has been assumed that the system starts from rest and β = /ω = 12 T/t1 . The response to the second sine wave is similar to that given by Equation 7.49 with t replaced by t − t1 . The total response in the era t > t1 is thus obtained from u(t) =

P0 1 {(sin t − β sin ωt) + sin (t − t1 )− β sin ω(t − t1 )} t > t1 k 1 − β2 (7.50)

The histories of displacement response obtained from Equations 7.49 and 7.50 for six different values of t1 /T are shown in Figure 7.16. The static displacement produced by the sinusoidal pulse is also shown in that figure by dashed lines. It may be noted that for t1 /T = 0.5, which corresponds to β = 1, Equations 7.49 and 7.50 both become indeterminate. The limiting values of these equations can however be determined by applying L’Hospital’s rule. Thus on differentiating the numerator and denominator of Equation 7.49 with respect to β and setting β = 1 we get u(t) =

1 P0 (sin ωt − ωt cos ωt) t ≤ t1 2 k

(7.51)

which is identical to Equation 6.17. For t1 /T = 0.5 the natural frequency ω = π/t1 , and Equation 7.51 reduces to   πt πt πt 1 P0 sin t ≤ t1 − cos (7.52) u(t) = 2 k t1 t1 t1

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Dynamics of structures

Figure 7.16 Time histories of response to a half-sine pulse load.

In a similar manner Equation 7.50 gives u(t) =

1 P0 (sin ωt − ωt cos ωt + sin ω(t − t1 )− ω(t − t1 ) cos ω(t − t1 )) t > t1 2 k (7.53)

On substituting ω = π/t1 Equation 7.53 reduces to π πt u(t) = − cos 2 t1

t > t1

(7.54)

Referring to Figure 7.16 when t1 /T = 0.25 the maximum response occurs in the free-vibration era and the maximum value is 0.943 P0 /k. For t1 /T = 0.5 the maximum response occurs at t = t1 , that is at the end of the pulse and the maximum value is 1.414 P0 /k. For t1 /T = 1.0 and 1.5 there is a single peak in the forced-vibration era; for t1 /T = 2.0 there are two peaks, the first one of which is larger; and for t1 /T = 3.0 there are three peaks, in which the second is the largest. For the special case of t1 /T = (2n + 1)/2 both the displacement and velocity are zero at the end of the pulse, as can be verified from Equation 7.49 and its differential, and the motion ceases at the end of the pulse.

Response to general dynamic loading and transient response

303

In order to obtain the shock spectrum we need to determine whether the maximum occurs in the forced-vibration era or the free-vibration era. If the maximum displacement occurs in the forced-vibration era t < t1 , the time to maximum, tp , is obtained by equating the differential of Equation 7.49 to zero. This leads to cos tp = cos ωtp

(7.55a)

and hence to tp = 2π n ± ωtp

n = 0, 1, 2, . . .

(7.55b)

The positive sign in Equation 7.55b corresponds to local minima, while the negative sign corresponds to local maxima. The time at which maximum occurs is thus given by tp =

2nπ 2n = +ω 1/t1 + 2/T

n = 1, 2, 3, . . .

(7.56)

For the maximum to occur within the forced-vibration era tp must be less than or equal to t1 giving 2n ≤ t1 1/t1 + 2/T

(7.57a)

2n − 1 t1 ≥ T 2

(7.57b)

or

The smallest value of t1 for which the maximum will occur in the forced-vibration era is obtained by substituting n = 1 in Equation 7.57b 1 t1 ≥ T 2

(7.58)

Since t1 /T = 1/2β, Equation 7.58 is equivalent to the condition β ≤ 1. The maximum response is obtained by substituting for t = tp in Equation 7.49. umax 1 = P0 /k 1 − β2




2πnβ 2πn sin − βsin 1+β 1+β

 (7.59)

A peak response is obtained corresponding to each value of n in Equation 7.59. However the maximum value of n is limited by the need to satisfy Equation 7.57b. Of the several peaks that may be obtained in the forced-vibration era the one that gives the largest response is used in determining the shock spectrum.

304

Dynamics of structures

For t1 /T < 1/2 or β > 1, the maximum response occurs in the free-vibration era t > t1 , for which Equation 7.50 holds. By substituting t1 = π/ , Equation 7.50 can be simplified to
 P0 2β π π u=− cos sin ωt − (7.60) k 1 − β2 2β 2β Equation 7.60 represents a harmonic motion with amplitude u=

P0 2β π cos k 1 − β2 2β

Hence the maximum response ratio is given by umax π 2β cos = 2 1−β 2β P0 /k

1 t1 < T 2

(7.61)

The shock spectrum for the sine pulse obtained from Equations 7.59 and 7.61 is plotted in Figure 7.17.

7.7.4 Effect of viscous damping As stated earlier, from the point of view of design the response parameter of particular interest is the maximum displacement. For a short-duration load the maximum displacement is not significantly affected by the damping normally present in a structure. As will be evident from the displacement histories shown in Figures 7.10, 7.13, and 7.16, for a short-duration pulse the maximum displacement is usually reached after

Figure 7.17 Shock spectrum for a sinusoidal pulse load.

Response to general dynamic loading and transient response

305

about one-fourth of a cycle of motion. For such a short duration of motion the amount of energy that can be dissipated through damping is quite small. As an example of the effect of damping consider the response of a damped system to a rectangular pulse function of duration t1 . Equation 7.16 shows that if t1 ≥ π/ωd , the maximum displacement occurs in the forced-vibration era and its value is given by Equation 7.17. When t1 < π/ωd , maximum displacement is in the free-vibration era, and to determine its value we need the displacement and velocity at t = t1 . These are obtained from Equation 7.14 and its derivative. 
 P0 ωξ u0 (t1 ) = 1 − e−ωξ t1 cos ωd t1 + (7.62) sin ωd t1 k ωd v0 (t1 ) P0 1 = e−ωξ t1 sin ωd t1 ωd k 1 − ξ2

(7.63)

The response in the free-vibration era is obtained from Equation 5.38 u(t) = ρe−ωξ t¯ sin(ωd t¯ + φ)

(7.64)

where t¯ = t − t1 and ρ and φ are given by Equation 5.39. The time at which a peak is reached in the displacement response is obtained from Equation 5.41, so that sin(ωd t¯ + φ) =



1 − ξ2

(7.65)

The peak value of the displacement is determined by substituting in Equation 7.64 the value of t¯ obtained from Equation 7.65. The shock spectrum obtained by the method described in the previous paragraphs is plotted in Figure 7.18 for five different values of the damping ratio ξ . It will be

Figure 7.18 Shock spectra for a rectangular pulse load corresponding to different values of the damping ratio.

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Dynamics of structures

observed that for t1 /T = 0.5 the maximum displacement reduces by only about 13.5% as the damping changes from 0 to 10%.

7.7.5 Approximate response analysis for short-duration pulses For a short-duration pulse it possible to obtain a reasonably good estimate of the maximum response by assuming that the force pulse is concentrated at t = 0 and has an impulse whose magnitude is equal to the area enclosed by the force pulse under consideration. For the three types of force pulses discussed in the earlier paragraphs the magnitude of impulse I is given by For rectangular pulse: I = P0 t1 For triangular pulse: I = 12 P0 t1 For sinusoidal pulse: I = 2P0 t1 /π where P0 is the amplitude of each of the three pulses. The response produced by an impulse I imparted at time t = 0 is obtained from I sin ωt mω Iω = sin ωt k

u=

(7.66) (7.67)

The maximum displacement is thus Iω/k. The approximate value of the displacement spectrum calculated from Equation 7.66 is compared in Figure 7.19 with the more exact values presented in earlier sections. For a relatively short duration of the pulse, for example t1 /T < 0.25, the approximate analysis provides a reasonably good estimate. For t1 /T > 0.5, when the

Figure 7.19 Approximate and exact values of the displacement spectra for rectangular, triangular, and sinusoidal pulse loads.

Response to general dynamic loading and transient response

307

maximum is achieved in the forced-vibration era, Equation 7.66 is meaningless because it is based on the assumption that the peak response occurs in the free-vibration era.

7.8

RESPONSE TO GROUND MOTION

The analysis of dynamic response of a structure to ground motion is of considerable interest because of its application in the design for earthquake forces. Earthquake ground motions are quite random in nature. However if the time history of ground motion is available, it is possible to determine the response of a single-degreeof-freedom system subjected to such a motion. The equation of motion of an underdamped single-degree-of system subjected to ground motion is given by mu¨ t + cu˙ + ku = 0

(7.67a)

mu¨ + cu˙ + ku = −mu¨ g

(7.67b)

or

where ut is the absolute motion measured from a fixed frame of reference, and u is the displacement relative to the ground, so that ut = u + ug . On dividing both sides of Equation 7.67 by m we get u¨ + 2ξ ωu˙ + ω2 u = −u¨ g

(7.68)

Equation 7.67b is an equation of forced-motion where the forcing function p(t) = −mu¨ g . It can be solved for u by a numerical evaluation of the Duhamel’s integral or by direct numerical integration using methods described in Chapter 8. It will be observed that for a given ground motion the response of the single-degree-of-freedom system depends only on the frequency ω and the damping ratio ξ . Therefore once a ground motion has been selected we can construct a series of curves that provide the maximum response of a system to the selected ground motion for a range of values of the system natural frequency and damping. Usually each curve in the series is constructed for one specific value of the damping and plots the maximum displacement versus the frequency, or the period. As stated earlier, such plots are referred to as response spectrum curves. Once the spectra have been determined, the maximum displacement and forces induced in a single-degree-of-freedom system having specified dynamic characteristics by the given earthquake ground motion are easily calculated. Consider the Duhamel integral solution of a single-degree-of-freedom system subjected to a ground acceleration u¨ g u(ω, ξ ) =

1 mωd



t

−mu¨ g (τ )e−ξ ω(t−τ ) sin ωd (t − τ ) dτ

(7.69)

0

The maximum value of u obtained from Equation 7.69 is referred to as the spectraldisplacement and denoted by Sd . Thus Sd (ω, ξ ) = umax (ω, ξ )

(7.70)

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Dynamics of structures

We will find it useful to define two related spectral parameters, referred to as the pseudo-spectral-velocity, Sv , and pseudo-spectral-acceleration, Sa . The two spectral parameters are given by Sv = ωSd

(7.71a)

Sa = ωSv = ω2 Sd

(7.71b)

The pseudo-spectral-velocity is not the same as the true spectral velocity. The latter can be obtained by evaluating the differential of Equation 7.69 and taking the maximum value of the differential. Differentiation under the integral sign gives ξω u(t) ˙ =− u+ ωd



t

u¨ g (τ )e−ξ ω(t−τ ) cos ωd (t − τ ) dτ

(7.72)

0

The true spectral velocity V = u˙ max is determined by taking the maximum value of u˙ obtained from Equation 7.72. The pseudo-spectral-acceleration is approximately equal to the maximum value of the true absolute acceleration. The latter can be obtained from the equation of motion. Thus u¨ t (t) = −2ξ ωu(t) ˙ − ω2 u(t)

(7.73)

For low values of damping u¨ t (t) ≈ −ω2 u(t), so that u(t) and u¨ t (t) attain their maximum values at the same instant of time and u¨ tmax ≈ ω2 umax = ω2 Sd = Sa . For an undamped system the pseudo-spectral acceleration and the maximum absolute acceleration are identical; for low amounts of damping the two are nearly equal. The pseudo-spectral-velocity and pseudo-spectral-acceleration have useful physical properties. The pseudo-spectral velocity provides a measure of the peak value of strain energy Es(max) stored in the system. Es(max)


2 1 2 1 2 1 1 Sv = kumax = kSd = k = mSv2 2 2 2 ω 2

(7.74)

The pseudo-spectral acceleration can be used to determine the peak value of the spring force. fS(max) = kumax = mω2 Sd = mSa

(7.75)

Because of the relationship between Sd , Sv , and Sa given by Equation 7.71, all three of them can be plotted on a single four-way logarithmic graph. In such a plot log Sv is plotted on the ordinate against log T (or log ω) on the abscissae. Equation 7.71a gives Sv =

2π Sd T

log Sv + log T = log 2π + log Sd

(7.76)

Response to general dynamic loading and transient response

309

For a constant Sd Equation 7.76 represents a straight line on the four-way log graph sloping to the left with a slope of −1. Lines sloping to the left with a slope of −1 thus represent constant Sd lines. A line perpendicular to these lines provides the log scale for the spectral-displacement. In a similar manner Equation 7.71b, gives Sa =

2π Sv T

log Sv − log T = log Sa − log 2π

(7.77)

For a constant Sa Equation 7.77 represents a straight line on the four-way log graph sloping to the right with a slope of 1. Lines sloping to the right with a slope of 1 thus represent constant Sa lines and a line perpendicular to these lines provides a log scale for the pseudo-spectral-acceleration. As an example of the application of response spectra, consider the one-story shear frame shown in Figure 7.20a. The columns are considered to be axially rigid and of negligible mass. The floor is infinitely stiff and all of the mass is concentrated at the floor level. The system has only one degree of freedom; the lateral displacement at the floor level. Damping is of viscous type and is represented by the dash pot shown on the diagram. The floor mass is m, the total stiffness of the columns is k andthe damping constant is c. Correspondingly, the natural period of vibration is T = 2π m/k and the damping ratio is ξ . These two parameters are varied over a range of values. The frame is subjected to a lateral ground motion. The variation of ground acceleration u¨ g with time is shown in Figure 7.20b, which also shows the histories of ground velocity and displacement. The maximum value of the lateral displacement of mass m relative to the ground will be attained either during the time the ground motion lasts or within at most half a cycle of free vibration after the ground ceases to move. For a selected value of the period T and damping ratio ξ the history of ground displacement is obtained for the duration of ground motion plus a half-cycle of free vibration by using one of the numerical methods of integration described in Chapter 8. In the present case the method that relies on a piece-wise linear representation of the excitation is most effective. The spectral-displacement Sd , that is the maximum value of relative displacement, corresponding to the given period and damping is obtained from the time history of motion. The pseudo-spectral-velocity and acceleration are now determined from Equation 7.71. The spectral-velocity has been plotted against the natural period for two values of damping, ξ = 0.0 and 0.1, on a four-way logarithmic graph in Figure 7.21. The displacement and acceleration scales are shown in Figure 7.21 along lines sloping to the right and left, respectively. Several properties of interest can be observed from the response spectrum plots of Figure 7.21. For very short period structures the stiffness k would be very large, and the mass m would more-or-less move along with the ground. The absolute acceleration of the mass should therefore be close to that of the ground. This can be verified from Figure 7.21 where the maximum value of ground acceleration has been shown by dashed line. For a long period structure k approaches zero and the structure is so flexible that the mass remains stationary even as the ground moves. The relative displacement of the mass should therefore be equal to the ground displacement. This is

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Dynamics of structures

Figure 7.20 (a) Single-story shear frame; (b) variation of ground motion with time.

seen to be the case from Figure 7.21 where the maximum value of ground displacement has also been shown. It would be of interest to compare the relative values of the true spectral-velocity and the pseudo-spectral-velocity. This has been done in Figure 7.22 for ξ = 0.1. In the intermediate period range the two are quite close. In the long period range the true spectral-velocity is greater than the pseudo-spectral velocity. This would be expected

Figure 7.21 Variation of spectral velocity with natural period.

Figure 7.22 Comparison between the true spectral velocity and pseudo-spectral velocity.

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Dynamics of structures

Figure 7.23 Comparison between the true spectral acceleration and pseudo-spectral acceleration.

for the following reason. For a very flexible structure the relative displacement and relative velocity of the mass m are equal to the ground displacement and ground velocity, respectively. Thus, Sd = ug0 and V = u˙ g0 where V is the true-spectral-velocity and ug0 , u˙ g0 are the maximum ground displacement and velocity, respectively. On the other hand, the pseudo-spectral-velocity approaches zero when T approaches ∞ as would be seen from Equation 7.71a. In the short period range the true spectral-velocity is smaller than the pseudo-spectral velocity. The pseudo-spectral-acceleration has been compared with the true spectralacceleration in Figure 7.23 for a damping ratio ξ = 0.1. As would be expected, the two are quite close. To illustrate the application of response spectrum in design, suppose that it is required to determine the maximum relative displacement as well as the maximum base sear induced in the frame of Figure 7.20a by the design ground motion shown in Figure 7.20b. The natural period of the frame is given as 1.0 s and the damping ratio ξ = 0.1. The Sd and Sa values for a period of 1 s and damping ratio 0.1 can be read off from the response spectrum shown in Figure 7.21. From the numerical data used to plot the spectral curves in Figure 7.21 the following values are obtained. Sd = 0.435 m,

Sa = 17.16 m/s2

The maximum relative displacement is thus 0.435 m and the maximum base shear is Vb = kSd = mω2 Sd = m(2π/T)2 Sd = 17.16m. The base shear can also be obtained from the pseudo-spectral acceleration. Thus Vb = mSa = 17.16m.

Response to general dynamic loading and transient response

313

7.8.1 Response to a short-duration ground motion pulse As stated earlier, damping can reasonably be neglected in determining the response to a short-duration pulse. Neglecting damping the equation of motion for ground excitation specializes to mu¨ t + ku = 0

(7.78a)

mu¨ + ku = −mu¨ g

(7.78b)

or

Equation 7.78b is an equation of forced-vibration response in which the effective force is given by p = −mu¨ g . When maximum displacement is the quantity of interest, the negative sign in the forcing function is of no consequence, and if u¨ g0 represents the maximum amplitude of ground acceleration, the amplitude of effective force is given by P0 = mu¨ g0 . The response ratio Rd is then obtained from Rd =

ω2 umax umax = mu¨ go /k u¨ go

(7.79)

Equation 7.79 shows that the shock spectra derived earlier for different types of force impulses also apply to a ground motion pulse when the ordinate is taken to be equal to ω2 umax /u¨ g0 . This statement is, however, true only when the initial conditions represented by u(0) and u(0) ˙ are both zero. From Equation 7.78a we note that u¨ tmax = −ω2 umax , hence the shock spectrum also gives the absolute value of u¨ tmax /u¨ g0 . Example 7.3 The base of an undamped single-degree-of-freedom system is subjected to a velocity pulse shown in Figure E7.3a. Obtain the response of the system and plot the shock spectrum of relative motion.

Solution The velocity pulse function is given by
 t u˙ g = v0 1 − t1

(a)

from which the ground acceleration is obtained as u¨ g = −

v0 t1

(b)

The equation of motion, written in terms of the relative displacement, is mu¨ + ku =

mv0 t1

(c)

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Dynamics of structures

Figure E7.3 Shock spectrum for ground velocity pulse: (a) ground velocity pulse; (b) shock spectrum. with the initial condition u0 = 0 and u˙ 0 = u˙ t − u˙ g0 = −v0 . Using Duhamel’s integral and the expression for free-vibration response, the total response is obtained as follows: u=−

v0 1 sin ωt + ω mω



t

o

mv0 sin ω(t − τ ) dτ t1

t ≤ t1

(d)

and u=−

1 v0 sin ωt + ω mω



t1

o

mv0 sin ω(t − τ ) dτ t1

t > t1

(e)

Evaluating the integrals in Equations d and e, we obtain v 1 0    ω −sin ωt + u=  v 1   0 −sin ωt + ω

1 (1 ωt1

2

− cos ωt)

1 {cos ω(t ωt1

2

− t1 ) − cos ωt}

t ≤ t1

(f)

t > t1

(g)

Response to general dynamic loading and transient response

315

If the peak response occurs for t < t1 , the time to maximum response is obtained by equating the differential of Equation f to zero. This gives tan ωtp = ωt1

(h)

Substitution of t = tp in Equation f gives the shock spectrum 1 1 ωt1 1 umax ω = ± ±  2 v0 ωt1 ωt 1 1 + (ωt1 ) 1 + (ωt1 )2

(i)

If the maximum response occurs in the free-vibration era, t > t1 , its value should be obtained from Equation g, which can be expressed as uω = ρsin(ωt + φ) v0

(j)

where  ρ = 1+

2 2 − (ωt1 )2 ωt1


sin ωt1 +

cos ωt1 ωt1

1/2

and tan φ =

cos ωt1 − 1 sin ωt1 − ωt1

The maximum response is equal to the amplitude ρ. The shock spectrum obtained from Equations i and j is plotted in Figure E7.3b. For t1 /T < 0.715, the maximum occurs in the freevibration era, t > t1 ; for t1 /T > 0.715, the maximum occurs during t < t1 . At t1 /T = 0.715, the maximum value of response uω/v0 is 1.247.

7.9 ANALYSIS OF RESPONSE BY THE PHASE PLANE DIAGRAM Let an undamped single-degree-of-freedom system be subjected to a base displacement pulse of magnitude ug and duration t1 , as shown in Figure 7.24a. From Section 5.2.1 we know that the total response of such a system can be obtained from a phase plane diagram. As shown in Figure 7.24b, the phase vector rotates with its center located a distance ug from the origin. If the system starts from at rest condition, the tip of the vector is at the origin. In the era of motion from 0 to t1 , the vector rotates through an angle ωt1 to position CA. If the base now returns to its original position, the subsequent era of motion is represented by vector OA rotating about O. The displacement and velocity at A constitute the initial conditions for the second phase of motion. The displacement–time relationship for both eras of motion is shown in Figure 7.24c. The equation of motion for the system subjected to a base displacement impulse is given by mu¨ t + kut = kug0

t ≤ t1

mu¨ + ku = 0

t > t1

t

t

(7.80)

316

Dynamics of structures

Figure 7.24 Phase plane diagram for a ground displacement pulse.

Figure 7.25 Response to an arbitrary load: (a) loading; (b) phase plane diagram for an undamped system.

The equation of motion of a system subjected to a rectangular pulse of magnitude P0 and duration t1 is mu¨ + ku = P0

t ≤ t1

u¨ + ku = 0

t > t1

(7.81)

On comparing Equations 7.80 and 7.81 and noting that for no base motion u ≡ ut , we observe that the motion caused by a rectangular pulse is similar to that caused by a base displacement, provided that kug0 is replaced by P0 . This suggests that a phase plane diagram may also be effectively utilized to obtain the response to a rectangular pulse force. The center of rotation of the phase vector will, in this case, be located at a distance P0 /k from the origin. In all other respects, the phase plane diagram will be similar to that for base displacement impulse. The phase plane method can also be extended to the analysis of response of a general dynamic loading. The loading is divided into a series of rectangular pulses of short durations as shown in Figure 7.25a. The center of rotation of the phase vector

Response to general dynamic loading and transient response

317

corresponding to the nth pulse, of magnitude Pn , will be located a distance Pn /k from the origin. The initial conditions for the phase of motion from tn−1 to tn will be those existing at the end of the previous pulse. The resulting phase plane diagram will be as shown in Figure 7.25b. In this diagram, the applied load has been represented by three force pulses of heights P1 , P2 , and P3 and durations t1 , t2 , and t3 , respectively. The system is assumed to be undamped and initially at rest. In the first phase of response, the center of rotation O1 is located at a distance P1 /k from the origin, and vector O1 A rotates at ω rad/s through an angle ωt1 to O1 A1 . In the second phase of response, the center of rotation is located at O2 , a distance P2 /k from the origin, and the starting position of the vector is O2 A1 . At the end of this phase, the vector has rotated by an angle ωt2 to position O2 A2 . In the third phase of response, vector O3 A2 rotates to position O3 A3 , where AO3 = P3 /k and angle A2 O3 A3 = ωt3 . The displacement u and the velocity function u/ω ˙ can now be obtained by taking the projections of the rotating vector on the vertical and horizontal axis, respectively. SELECTED READINGS Ayre, R.S., “Transient Response to Step and Pulse Functions’’, in Shock and Vibration Handbook, 4th Edition, (ed. C.M. Harris), McGraw-Hill, New York, 1996. Bishop, R.E.D., Parkinson, A.G., and Pendered, J.W., “Linear Analysis of Transient Vibration’’, Journal of Sound and Vibration, Vol. 9, 1969, pp. 313–337. Crede, C.E., Vibration and Shock Isolation, John Wiley, New York, 1951. Jacobsen, L.S. and Ayre, R.S., Engineering Vibrations with applications to structures and machinery, McGraw-Hill, New York, 1958. Matsuzaki, Y., and Kibe, S., “Shock and Seismic Response Spectra in Design Problems, Shock and Vibration Digest, Vol. 15, 1983, pp. 3–10. Thomson, W.T., Theory of Vibration with Applications, 5th Edition, Prentice Hall, Upper Saddle River, NJ, 1998.

PROBLEMS 7.1

A single degree-of-freedom system with mass m, stiffness k and negligible damping is subjected to alternating step force shown in Figure P7.1. Note that the applied force is

Figure P7.1

318

Dynamics of structures periodic with a period equal to the natural period of the system. Plot the response of the system as a function of time. Show that resonance takes place and that the successive peaks of response have a magnitude of 2nP0 /k where n is the number of half cycles from the starting point.

7.2

Obtain the response of the system shown in Figure P7.2 to an applied force F which varies as indicated.

Figure P7.2

7.3

A single degree-of-freedom system is subjected to shock loading shown in Figure P7.3. Determine the condition under which the maximum response will be obtained for t > 3t1 and find the magnitude of such response.

Figure P7.3

7.4

In Problem 7.3, assume that t1 /T is very small so that the applied force can reasonably be viewed as an impulsive force applied at t = 0. Obtain an expression for the resulting response for t > 3t1 and hence find the maximum response. If t1 /T = 1/32, how much is the approximate response in error as compared to the exact value of the maximum obtained in Problem 7.3.

7.5

A ground velocity pulse v0 applied at time t = 0 can be interpreted as a sudden change in ground velocity at that time (Fig. P7.5). Show that such a change in velocity will be caused by an acceleration input of v0 δ(t), where δ(t) is the delta function. Using Duhamel’s

Response to general dynamic loading and transient response

319

integral and the properties of the delta function, prove that the response of the system to the velocity pulse is given by u=−

v0 −ξ ωt e sin ωd t ωd

Hence show that the displacement response spectrum is obtained from ! "  ξ v0 umax = exp −  sin−1 1 − ξ 2 ω 1 − ξ2 Plot the response spectra (umax as a function of period T) for ξ = 0, 0.05 and 0.1.

Figure P7.5 7.6

A single degree-of-freedom system is subjected to a force which varies as shown in Figure P7.6. Obtain a plot of the maximum value of response as a function of t1 /T, where T is the natural period of the system.

Figure P7.6 What will be the maximum displacement of the system when P0 = 10 kN, and t1 = 1/2 s. The natural period of the system is 1s and the spring stiffness is 250 kN/m.

320 7.7

Dynamics of structures Obtain the response of a single degree-of-freedom system subjected to the ground velocity pulse shown in Figure P7.7.

Figure P7.7 7.8

Obtain expressions for the response to step function load with rise time shown in Figure 7.7 by the Duhamel’s integral.

7.9

Using Duhamel’s integral, obtain expressions for the response to the triangular pulse force shown in Figure 7.12.

7.10

Obtain the response of an undamped single-degree-of-freedom system to the double rectangular pulse shown in Figure P7.10 for time t > 2t1 + t0 . The duration of each pulse is t1 and the time lapse between the two pulses is t0 . Determine the value of response when t0 = (π/ω) − t1 .

Figure P7.10 7.11

An automobile modeled as a single-degree-of-freedom system rides over a hump in a road. The hump can be represented by a half sine wave of length L and height h as shown in Figure P7.11. The speed of the vehicle is v and its natural frequency is ω. Assuming that the shock absorbers of the car are worn, so that the damping is zero, show that the absolute displacement of the vehicle while it is riding over the hump is given by: ut =

h (sint − β sin ωt) 1 − β2

where  = π v/L and β = /ω. Also show that the displacement of the vehicle relative to the ground while the vehicle is riding over the hump is given by: u=

hβ 2 1 − β2


sin t −

 1 sin ωt . β

Response to general dynamic loading and transient response

321

Figure P7.11 7.12

The following data are supplied for the vehicle and the road referred to in Problem 7.11. L = 1.5 m, h = 50 mm, v = 20 km/hr. The mass of the vehicle is 1400 kg, and the stiffness of its spring is 40,000 N/m. Find the absolute maximum displacement of the vehicle and the maximum spring force. What is the maximum acceleration experienced by the passengers?

7.13

Repeat Problem 7.12 assuming that the damping is 40% of critical.

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Chapter 8

Analysis of single-degree-of-freedom systems: Approximate and numerical methods

8.1

INTRODUCTION

In many problems of the vibration of structural or mechanical systems, the quantity of primary interest is the predominant frequency of vibration. When the system has only one degree of freedom, the frequency of vibration is easily determined by the methods described in earlier chapters. Very often, however, the system under consideration is in fact a multi-degree-of-freedom system. Frequency determination of such systems is complex and time consuming. On the other hand, if a reasonable estimate can be made of the vibration shape, the system can be represented by an equivalent single-degreeof-freedom model whose frequency of vibration is more easily determined. Because we cannot expect the estimated vibration shape to be the same as the true vibration shape, the frequency determined by this method is in error. The practical utility of the method, however, lies in the fact that even with a crude estimate of the vibration shape, the error in the calculated frequency is generally small enough so that the frequency estimate is acceptable for all practical purposes. The frequency of a single-degree-of-freedom system, whether real or idealized, is most conveniently determined by the principle of conservation of energy. The procedure is commonly known as the Rayleigh method after Lord Rayleigh, who first proposed it in his book on the theory of sound. Its main application is to a system that can be idealized as a single-degree-of-freedom system by assuming an approximate vibration shape. Rayleigh showed that the frequency obtained by such an assumption is always higher than the true frequency. In this chapter we present a detailed description of the application of Rayleigh method in estimating the approximate frequency of an idealized single-degree-of-freedom system. In many instances, the complete history of vibration or at least the maximum displacement amplitude attained during an era of vibration, rather than just the frequency, is of interest. The structural or mechanical system can be set vibrating whenever it is disturbed from its position of rest by being subjected to an initial displacement or an initial velocity or both. The system may also be excited by the application of a dynamic force. Often the source of excitation will be a combination of initial disturbances and an applied force. In each case a response history calculation requires the solution of the equation of motion. Consider first a linear system. When such a system is excited by initial disturbances but is not subjected to any external force, closed-form mathematical solution can be

324

Dynamics of structures

obtained for the equation of motion. These solutions have been presented in the earlier chapters of the book. Thus, for a damped single-degree-of-freedom system subjected to an initial displacement uo and an initial velocity vo , the displacement response is given by

u(t) = e−ξ ωt




vo + ξ ωuo sin ωd t + u0 cos ωd t ωd

 (8.1)

A mathematical solution can also be obtained for the response of a linear system excited by an applied dynamic force provided that such a force can be represented by a simple mathematical function. The response of a linear system subjected to an exciting force p(t) is given by the Duhamel’s integral:

u(t) =

1 mωd



t

e−ξ ω(t−τ ) p(τ ) sin ωd (t − τ )dτ

(8.2)

o

Exact evaluation of the Duhamel’s integral is possible provided that p(t) is a simple function. Response solutions for several such exciting functions have been presented in Chapter 7. However, if p(t) is a complicated function or is defined only by numerical values at discrete intervals of time, mathematical evaluation of the integral in Equation 8.2 is not possible and a numerical method must be used to solve the equation of motion. The numerical methods used to find the response of a linear single-degree-offreedom system can be classified into the following categories: 1 2

Evaluation of the Duhamel’s integral by a numerical method Direct numerical integration of the equation of motion

A numerical evaluation of the Duhamel’s integral provides the response of the system to an applied force. If the system starts with prescribed initial conditions, the response due to the specified initial displacement and velocity must be obtained from Equation 8.1 and superimposed on the response obtained by the numerical evaluation of the Duhamel’s integral. On the other hand, when a direct numerical integration of the equation of motion is used, the combined response due to the initial conditions as well as the applied force can be directly obtained. For a nonlinear system the Duhamel integral solution is no longer valid and recourse must be had to a direct numerical integration of the equation of motion. In this chapter we present several alternative methods for the numerical evaluation of the Duhamel’s integral. This is followed by a detailed description of some of the methods used for the direct numerical integration of the equations of motion. It is observed that direct integration is a more general method since it is applicable to both linear and nonlinear systems. Also, as will be shown later, direct integration methods can be extended directly to the response analysis of a multi-degree-of-freedom system.

Analysis of single-degree-of-freedom systems 325

8.2

CONSERVATION OF ENERGY

Consider the undamped single-degree-of-freedom system shown in Figure 8.1a. The system is vibrating without the application of an external force. The equation of motion for such a system is given by mu¨ + ku = 0

(8.3)

The displacement and velocity responses of the system are shown in Figure 8.1b and c, respectively. On multiplying Equation 8.3 by velocity u˙ and integrating, we obtain 1 1 m(u) ˙ 2 + ku2 = C 2 2

(8.4)

where C is a constant. The first term on the left-hand side of Equation 8.4 represents the kinetic energy of the system, while the second term represents the potential energy, which in this case is the strain energy of the spring. Equation 8.4 shows that for the system under consideration, the total energy, kinetic plus potential, is a constant. In general, for a system subjected only to conservative forces, the total energy is always conserved. A formal proof of the statement was provided in Section 4.9. In writing the energy equation, we must include in the total potential energy the potential of the applied force as well. Equation 8.4 thus represents a special case when no external force is acting on the system and the potential energy consists only of the internal strain energy.

Figure 8.1 Undamped single-degree-of-freedom system.

326

Dynamics of structures

We proved in Chapter 5 that a solution of Equation 8.3 is given by u = A sin(ωt + φ)

(8.5)

where ω is the natural frequency of the system and A and φ are arbitrary constants whose values are determined by the initial conditions. By using Equation 8.5 the kinetic energy T and the potential energy V can be shown to be 1 mω2 A2 cos2 (ωt + φ) 2 1 V = kA2 sin2 (ωt + φ) 2 T=

(8.6a) (8.6b)

For the kinetic energy to be a maximum, cos(ωt + φ) must be equal to 1. When such is the case, sin(ωt + φ), and therefore the potential energy, is zero. At such an instance the total energy is equal to the kinetic energy, given by Tmax =

1 mω2 A2 2

(8.7)

In a similar manner, the potential energy is a maximum when sin(ωt + φ) is 1, at which time cos(ωt + φ) and therefore the kinetic energy is zero. At such an instant, the total energy is equal to the potential energy given by Vmax =

1 2 kA 2

(8.8)

Because the total energy is constant, using Equations 8.4, 8.7, and 8.8, we obtain C = Tmax = Vmax

(8.9a)

1 1 mω2 A2 = kA2 2 2

(8.9b)

or

Equation 8.9 enables us to find an expression for the frequency ω: ) ω=

k m

(8.10)

The procedure just outlined, which uses the principle of conservation of energy, is known as Rayleigh’s method. Equation 8.10 is, in fact, identical to the one derived in Chapter 5 and gives the true vibration frequency of the system. It appears, then, that for the system considered, the Rayleigh method leads to the true frequency and seems to have no particular advantage over the direct method. This is in general true, although some problems are more easily solved by the energy method, as the following examples illustrate.

Analysis of single-degree-of-freedom systems 327

Example 8.1 A wheel and axle assembly of mass moment of inertia I0 is inclined from the vertical by an angle α, as shown in Figure E8.1. If an unbalanced weight w is attached to the wheel at a distance a from the axis and the assembly undergoes small vibrations about the position of rest, determine the frequency of vibration.

Figure E8.1 Vibration of a wheel axle assembly.

Solution If the air resistance and friction in the bearings are neglected, the only force acting on the system is that due to gravity. The gravity force being conservative, the total energy will be conserved. As the wheel swings, the height of the unbalanced weight above the datum and hence its potential energy changes. The kinetic energy, of course, depends on the angular velocity of the ˙ the kinetic energy T is wheel axle assembly. Denoting the angular velocity of the wheel by θ, given by T=

1 1w 2 2 2 I0 (θ˙ ) + a (θ˙ ) 2 2g

(a)

The height of the unbalanced weight changes from h1 to h as the wheel swings through an angle θ. Taking the datum as the lowest position of weight w, the potential energy V is given by V = w(h − h1 ) = wa sin α(1 − cos θ)

(b)

For small angle θ, Equation b becomes V = wa sin α

θ2 2

(c)

328

Dynamics of structures

For simple harmonic motion θ = A sin ωt

(d)

θ˙ = Aω cos ωt and Tmax = Vmax

(e)

From Equations a, c, and d, Tmax =

A2 ω 2 2

Vmax = w


 w I0 + a2 g

aA2 sin α 2

(f)

Finally, from Equations e and f, ω2 =

wa sin α I0 + (w/g)a2

(g)

Example 8.2 The system shown in Figure E8.2 is rotating about the vertical axis with a constant angular speed of  rads/s. The mass m is attached to the end of a light cantilever strip which has a moment of inertia I for flexure in the plane of the paper but is rigid in the plane of rotation. Obtain the natural frequency of vibration of mass m.

Figure E8.2 Wheel with an attached mass rotating at constant angular speed.

Solution When the system is at rest, the gravity force on mass m is balanced by the spring force exerted by the cantilever strip. If the displacement of the mass m is measured from its position at rest, we may neglect the gravity force in our formulation. The only other force acting on mass m is the centripetal force given by m2 R, where R is the radial distance of mass m from the axis of rotation.

Analysis of single-degree-of-freedom systems 329 Since a central force, defined as a force that is always directed toward or away from a fixed point, is conservative, the total energy in the system is conserved. The kinetic energy of the system is given by 1 m(u) ˙ 2+C 2

T=

(a)

where C is a constant that will account for the kinetic energy of rotation of the mass as well as the flywheel. The potential energy consists of the strain energy of the spring and the work done by the centripetal force. To obtain the former, we determine an equivalent force P which when applied at the end of the cantilever will cause a tip deflection of u: Pl 3 =u 3EI

(b)

The strain energy U is equal to the work done by the average force 12 P in undergoing deflection u: 3EI 1 Pu = 3 u2 2 2l

U=

(c)

To obtain the work done by the centripetal force, we first assume that for small displacements the force is constant and is given by F = m2 (l + r)

(d)

We also need the displacement of mass m in the horizontal direction. Referring to Figure E8.2, the horizontal displacement over a small distance dx is given by dξ = dx(1 − cos θ ) 1 2 θ dx 2 1 = (u )2 dx 2 =

(e)

The total displacement, obtained by integrating Equation e, is  ξ=

l

0

1 2 (u ) dx 2

(f)

Now for a tip force P, P u = EI 


 x2 lx − 2

(g)

Substitution of Equation g into Equation f gives ξ=

P2 l 5 15(EI)2

(h)

330

Dynamics of structures

On using Equation b, ξ can be expressed in terms of u as ξ=

3 u2 5 l

(i)

The work done by the centripetal force is W = m2 (l + r)

3 u2 5 l

(j)

and the total potential energy becomes V=

3 EI 2 3 u2 u + m2 (l + r) 3 2 l 5 l

(k)

Noting that for simple harmonic motion energy is conserved, the following expression is obtained for frequency ω ω2 =

3 (EI/l 3 ) 2

+ 35 (m2 /l)(l + r) m/2

or 
 r 6  2 1+ 1+ 5 ω0 l

 ω = 2

ω02

(l)

where ω02 = 3EI/ml 3 .

8.3 APPLICATION OF RAYLEIGH METHOD TO MULTI-DEGREE-OF-FREEDOM SYSTEMS As noted earlier, the main application of Rayleigh method is to the determination of an approximate value of the fundamental frequency of vibration of a multi-degreeof-freedom system. The method relies on the estimation of a vibration shape for the system so that the latter is converted to an equivalent single-degree-of-freedom system. The frequency of the equivalent system is then obtained by applying the principle of conservation of energy. For an illustration of the procedure, consider the system shown in Figure 8.2, consisting of a mass M attached to a uniform spring having stiffness k and total mass m. Because the spring possesses a mass, the system has, in fact, an infinite number of degrees of freedom. However, if we make the assumption that the displacement of the spring–mass system is given by u(x, t) = z(t)ψ(x)

(8.11)

where z(t) is a generalized coordinate, considered as the unknown, and ψ(x) is a shape function that must be selected, the system reduces to a single-degree-of-freedom system, because once z(t) is determined, the displaced shape of the entire system is known.

Analysis of single-degree-of-freedom systems 331

Figure 8.2 Vibration of mass M attached to a spring having mass m.

By definition, the spring constant k is the force required to stretch the spring of length l by a unit distance. Thus the force required to stretch an element of length dx by a unit distance will be lk/dx. Noting that the extension of the element is du, the force F acting on it is given by F=

l k du dx

(8.12)

The strain energy stored in the element is 1 du 1 F du = kl du 2 2 dx
2 1 du = kl dx 2 dx

dU =

(8.13)

The total strain energy is given by 1 U = kl 2

 l
0

1 = klz2 2



l

du dx

2 dx

(ψ  )2 dx

(8.14)

0

The total kinetic energy of the spring and the attached mass is readily found to be 

l

1 1 2 m(x){ ¯ u(x, ˙ t)}2 dx + M{u(l)} ˙ 2 2 o  1m 2 l 1 (˙z) = {ψ(x)}2 dx + M(˙z)2 {ψ(l)}2 2 l 2 o

T=

(8.15)

where m ¯ is the mass per unit length of the spring. Applying the principle of conservation of energy and assuming that z = A sin ωt, we obtain ω = 2

'l 1 kl o {ψ  (x)}2 dx 2 'l 1 (m/l) o {ψ(x)}2 dx + 12 M{ψ(l)}2 2

(8.16)

332

Dynamics of structures

If the displacement of the spring is assumed to vary linearly, ψ(x) = x/l. The selected shape function ψ satisfies the boundary condition of zero displacement at the fixed end, x = 0. The potential and kinetic energies are obtained from Equations 8.14 and 8.15, respectively U=

1 2 kz 2

(8.17a)

T=

m 2 1 M+ z˙ 2 3

(8.17b)

and

Assuming simple harmonic motion and equating the maximum values of U and T, we get % ω=

k M + m/3

(8.18)

Equation 8.18 shows that the frequency of the system of Figure 8.2 with a spring of mass m is approximately equal to that of an equivalent system with a massless spring and an attached mass that is equal to Mplus one-third the mass of the spring. When m is zero, Equation 8.18 reduces to ω = k/M, which is the true frequency of a system with massless spring. This is because the linear displacement shape is, in fact, the true vibration shape of such a system. As an alternative to the linear displacement shape, let us assume that ψ = sin(πx/2l). Note that, with this choice, too, u = 0 at x = 0 so that the boundary condition at the fixed end is satisfied. Substitution into Eqs 8.14 and 8.15 gives π2 1 U = klz2 2 2 4l =



l

cos2

0

πx dx 2l

2

1π kz2 2 8

(8.19)

and  1 1 m 2 l 2 πx sin (˙z) dx + M(˙z)2 2 l 2l 2 0 1m 2 1 = (˙z) + M(˙z)2 22 2

T=

(8.20)

The frequency ω is now given by π ω= √ 8

%

k M + 12 m

(8.21)

Analysis of single-degree-of-freedom systems 333

√  If the spring–mass m is zero,  Equation 8.21 will give ω = π/ 8 k/M. When compared to the true frequency of k/M this value is higher by about 11%. Onthe other hand, if the attached mass M is zero, the frequency works out to ω = π/2 k/m, which is exactly equal to the true frequency of vibration of a uniform spring. This is because sine function ψ = sin(πx/2l) happens to be the true vibration shape for that case. The foregoing example shows that the accuracy of the computed frequency depends on how closely the estimated vibration shape resembles the true vibration shape. Criteria for the selection of a vibration shape are discussed in a later section. The example also shows that even with a crude estimate of the vibration shape, the error in the calculated frequency is not too large. The application of Rayleigh method to discrete multi-degree-of-freedom systems follows lines similar to those described in the foregoing paragraphs. This is illustrated in the following example. Example 8.3 The three-story building frame shown in Figure E8.3 is modeled by assuming that all the mass is lumped at floor levels and that the columns are axially rigid. It is further assumed that the floor beams are infinitely rigid. This last assumption gives a model that is commonly referred to as a shear-type building model. The interstory stiffness of such a model can be easily obtained from known flexural rigidities of the columns. The story stiffness as well as the floor masses are shown in the figure. Determine the fundamental frequency of the building frame.

Figure E8.3 Lateral vibrations of a shear building: (a) building model; (b) displaced shape, ψ.

Solution The building model is, in fact, a three-degree-of-freedom system because the three displacement coordinates u1 , u2 , and u3 must be specified to determine the configuration of the frame. To convert the model to an equivalent single-degree-of-freedom system, we assume a suitable displacement shape ψ, where in this case ψ is a vector with three elements. In matrix notation we write u = z(t)ψ

(a)

334

Dynamics of structures

in which z(t) is the unknown generalized coordinate. The potential energy of the system is best expressed in matrix notation as 1 T u Ku 2 1 = {z(t)}2 ψ T Kψ 2

U=

(b)

where K is the stiffness matrix of the frame given by 

3.5 K = k −1.5 0

 −1.5 0 2.5 −1 −1 1

(c)

The kinetic energy of the frame is obtained from 1 T u˙ Mu˙ 2 1 = {˙z(t)}2 ψ T Mψ 2

T=

(d)

where M is the mass matrix of the frame given by  M = m



2.5



2

(e)

1 Let us now assume that the vibration shape is   1 ψ = 2 3

(f)

Substitution of Equations c and f into Equation b gives U=

1 2 z 4.5k 2

(g)

Also, from Equations d, e, and f we get T=

1 2 z˙ 19.5m 2

(h)

Assuming that the motion is simple harmonic and that energy is conserved, we get the following value of frequency from Equations g and h: ) ω = 0.48

k m

(i)

Analysis of single-degree-of-freedom systems 335

Figure 8.3 Flexural vibrations of a beam.

8.3.1 Flexural vibrations of a beam Consider the lateral vibrations of a simply supported nonuniform beam. The beam shown in Figure 8.3 has a mass of m(x) ¯ per unit length and flexural rigidity EI(x), both functions of coordinate x. The system represented by the beam has an infinite number of degrees of freedom. However, if we can make an estimate of the vibration shape, the system can be reduced to an equivalent single-degree-of-freedom system. Let us assume that deflection u(x, t) is given by u(x, t) = z(t)ψ(x)

(8.22)

where z(t) is the unknown coordinate and ψ(x) is the selected deflection shape. Standard textbooks on the mechanics of materials show that strain energy of the beam due to flexural deformations is given by 1 U = z2 2



L

EI(x){ψ  (x)}2 dx

(8.23)

o

The kinetic energy is obtained from 1 T = z˙ 2 2



L

2 m(x){ψ(x)} ¯ dx

(8.24)

o

The free undamped vibrations of the beam can be assumed as harmonic, so that z = A sin ωt. By substituting this value of z in Equations 8.23 and 8.24, we obtain the following expressions for the maximum potential and kinetic energies: Umax =

Tmax

1 2 A 2



L

EI(x){ψ  (x)}2 dx

(8.25)

0

1 = A2 ω 2 2



L

2 m(x){ψ(x)} ¯ dx

(8.26)

0

Applying the principle of conservation of energy, we get Tmax = Umax

(8.27)

336

Dynamics of structures

or 'L

EI(x){ψ  (x)}2 dx ω = '0 L 2 dx ¯ 0 m(x){ψ(x)} 2

(8.28)

Referring to Section 2.7.4, we note that the numerator in Equation 8.28 is identical to the generalized stiffness k∗ , while the denominator is equal to the generalized mass m∗ , so that ω2 =

k∗ m∗

(8.29)

It is observed that Rayleigh’s method gives the same result for the frequency as a formulation based on the concept of generalized coordinates. This is to be expected because the formulation is similar in both cases; z(t) in the Rayleigh method, being, in effect, a generalized coordinate. The only difference in the two approaches is that in the Rayleigh method, we have used the principle of conservation of energy, while in the generalized coordinate approach the virtual work principle was used. As a specific example, let us assume that the beam is uniform and that the shape function is ψ = sin(πx/L). Note that ψ is zero for both x = 0 and x = L. The assumed vibration shape therefore satisfies the two geometric boundary conditions. Substitution for ψ in Equations 8.23 and 8.24 gives U=

1 2 π4 z EI 3 4 L

(8.30)

T=

1 2 (˙z) mL ¯ 2

(8.31)

Assuming simple harmonic motion, and equating the maximum potential energy to the maximum kinetic energy, we get ) ω=π

2

EI mL3

(8.32)

where m = mL ¯ is the total mass of the beam. The frequency estimate given by Equation 8.32 is, in fact, exact because the selected shape function happens to be the true vibration shape of a uniform beam. Example 8.4 Find the fundamental frequency of the simply supported uniform beam shown in Figure E8.4a. Assume that the vibration shape of the beam is the same as the deflection shape assumed by the beam under (a) a uniformly distributed load (Fig. E8.4b), and (b) a central concentrated load (Fig. E8.4c).

Analysis of single-degree-of-freedom systems 337

Figure E8.4 Flexural vibrations of a uniform beam.

Solution (a) The deflected shape of the beam under a uniformly distributed load of w per unit length is given by y=

w EI




x4 Lx3 L3 x − + 24 12 24

 (a)

We now assume that the lateral deflection of the vibrating beam is given by u(x, t) = z(t)ψ(x)

(b)

where ψ(x) =

1 EI




x4 Lx3 L3 x − + 24 12 24

 (c)

and the arbitrary load w has been included in z. Substitution into Eqs 8.23 and 8.24 gives 1 2 1 L5 z 2 EI 120
2 1 31L9 1 T = z˙ 2 2 EI 362,880

U=

(d) (e)

338

Dynamics of structures

As an alternative, the strain energy can be obtained by recognizing that it is equal to the work done by the applied force, so that U= =

1 2



L

z · zψ(x)dx

0

1 2 L5 z 2 120EI

(f)

The vibration frequency is obtained from Equation 8.28: ) ω = 9.8767

EI mL3

(g)

where m is the total mass of the beam. In comparison with the exact frequency given by Equation 8.32, this is only 0.07% too high. (b) The deflected shape of the beam under a central concentrated load P is given by y=

P (3L2 x − 4x3 ) 48EI

for 0 ≤ x ≤ L/2

(h)

The deflections are symmetrical about the center line. Again, we assume that the lateral deflections are given by Equation b where ψ(x) =

1 (3L2 x − 4x3 ) 48EI

(i)

and the arbitrary load P has been included in z. The strain energy and the kinetic energy, respectively, are given by U= =

1 2 z 2 2



L/2

EI{ψ  (x)}2 dx

0

1 2 L3 z 2 48EI

(j)

and T= =

1 2 z˙ 2 2



L/2

2 m{ψ(x)} ¯ dx

0

1 2 mL ¯ 7 17 z˙ 2 (48EI)2 35

(k)

Applying Rayleigh’s principle, we obtain ) ω = 9.941

EI mL3

The frequency given by Equation l is 0.72% too high.

(l)

Analysis of single-degree-of-freedom systems 339 Note that the strain energy could be obtained by computing the work done by the external force, so that
 1 L U = z2 ψ 2 2 =

8.4

1 z 2 L3 2 EI 48

(m)

IMPROVED RAYLEIGH METHOD

The examples presented in the previous sections show that the frequencies obtained by assuming a vibration shape which is not the same as the true shape are always higher than the exact frequency. On physical considerations this can be explained by the fact that external constraints must be applied to the system if it has to be forced to vibrate in a shape that is different from its true vibration shape. These constraints make the system stiffer and hence increase its frequency. Later, we shall provide more rigorous mathematical proofs for the foregoing statement, both for discrete and continuous systems. The statement also implies that if by using the Rayleigh method, more than one estimate is obtained for the frequency of a system, the lowest of these estimated values is closest to the true frequency. It is also obvious that in the application of Rayleigh method, the closer the assumed vibration shape is to the true vibration shape, the better is the estimate of the frequency. We can systematically improve the assumption of vibration shape by recognizing that the dynamic deflections result from the application of inertia forces. For example, consider the free vibration equation of a discrete multi-degree-of-freedom system. Mu¨ + Ku = 0

(8.33)

which can be expressed in the alternative form u = −K−1 Mu¨

(8.34)

Equation 8.34 expresses the fact that the displacement u results from the application of inertia force Mu. ¨ If u is assumed to be equal to A sin(ωt + φ)ψ, Equation 8.34 reduces to ψ = ω2 K−1 Mψ

(8.35)

Equation 8.35 holds if ψ is the true vibration shape. However, if instead of ψ, we use an approximate vibration shape ψ 0 , we cannot expect Equation 8.35 to hold and the product on the right-hand side would give a new vector ψ¯ 1 which will be different from ψ 0 , the initial assumption of the vibration shape. ψ¯ 1 = ω2 K−1 Mψ 0

(8.36)

Intuitively, we can reason that ψ¯ 1 should be closer to the true vibration shape than ψ 0 is, so that if we use ψ¯ 1 in the Rayleigh method, we should obtain a better estimate

340

Dynamics of structures

of the frequency. The foregoing reasoning forms the basis of an improvement in the Rayleigh method. In practice, since ω is not known, we use inertia forces given by Mψ 0 rather than ω2 Mψ 0 . The resulting vibration shape, ψ 1 = K−1 Mψ 0 , is proportional to ψ¯ 1 . Thus ψ¯ 1 = ω2 ψ 1

(8.37)

Using ψ¯ 1 to obtain the strain energy, we get 1 2¯T ¯ A ψ 1 Kψ 1 2 1 = A2 ω4 ψ T1 Kψ 1 2 1 = A2 ω4 ψ T1 KK−1 Mψ 0 2 1 2 4 T = A ω ψ 1 Mψ 0 2

(U)max =

(8.38)

Equating this to the maximum kinetic energy given by Tmax = 12 A2 ω2 ψ T0 Mψ 0 , we get ω2 =

ψ T0 Mψ 0 ψ T1 Mψ 0

(8.39)

We may, however, get a further improvement in the frequency if we use ψ¯ 1 instead of ψ 0 in calculating the maximum kinetic energy. Thus 1 2 2¯T ¯ A ω ψ 1 Mψ 1 2 1 = A2 ω6 ψ T1 Mψ 1 2

Tmax =

(8.40)

Equations 8.38 and 8.40 give ω2 =

ψ T1 Mψ 0 ψ T1 Mψ 1

(8.41)

The estimated value can be improved even further by using ψ 1 to obtain a new vibration shape ψ 2 . If the process is repeated a sufficient number of times, ψ n will eventually converge to the true vibration shape and the frequency obtained by using ψ n will be the exact frequency. We will provide a formal proof of this statement in a later chapter. However, because the advantage of Rayleigh method lies in obtaining a quick but reasonably accurate estimate of the frequency, the extra work involved in carrying the improvement too far is not justified.

Analysis of single-degree-of-freedom systems 341

The reasoning of the foregoing paragraphs applies equally well to a continuous system. Assume that the vibration shape of the flexural beam, ψ(x), is the same as the deflection shape produced by an external load p(x). Deflection u(x, t) = zψ(x) will therefore be caused by a force zp(x) and the strain energy stored in the beam will be given by U=

1 2



L

zp(x)zψ(x) dx 0

1 = z2 2



L

p(x)ψ(x) dx

(8.42)

0

Correspondingly, the maximum strain energy will be 1 2 A 2

(U)max =



L

p(x)ψ(x) dx

(8.43)

0

On equating this to the maximum kinetic energy given by Equation 8.26, we obtain p(x) = ω2 m(x)ψ(x) ¯

(8.44)

Thus an applied load p(x) equal to the inertia force produces a deflected shape that is identical to the vibration shape of the beam. The statement will be true provided that ψ(x) is the true vibration shape. On the other hand, if the assumed vibration shape denoted by ψ0 (x) in not the same as the true vibration shape, the deflection produced ¯ by force ω2 m(x)ψ 0 (x) will not be ψ0 (x) but slightly different from it. Let us represent such a deflected shape by ψ¯ 1 (x). Again, intuitively we can state that ψ¯ 1 (x) will be closer to the true vibration shape than ψ0 (x) is. We cannot yet derive ψ¯ 1 (x) from ψ0 (x) because ω2 in Equation 8.44 is unknown. We therefore use ψ1 (x), the deflected shape ¯ 1 (x) are related resulting from the application of load m(x)ψ ¯ 0 (x). Shapes ψ1 (x) and ψ by the equation ψ¯ 1 (x) = ω2 ψ1 (x)

(8.45)

and the strain energy expression obtained from Equation 8.43 becomes (U)max

1 = A2 2 =



L 0

1 2 4 A ω 2

¯ ω2 m(x)ψ ¯ 0 (x)ψ1 (x) dx



L

m(x)ψ ¯ 0 (x)ψ1 (x) dx

(8.46)

0

Equating the maximum strain energy, Equation 8.46, to the maximum kinetic energy given by Equation 8.26 with ψ0 (x) substituted in place of ψ(x), we get 'L

2 m(x){ψ ¯ 0 (x)} dx ω = ' L0 ¯ 0 (x)ψ1 (x) dx 0 m(x)ψ 2

(8.47)

342

Dynamics of structures

We can, however, obtain a further improvement by using ψ¯ 1 (x) instead of ψ0 (x) in the kinetic energy expression Equation 8.26, so that Tmax =

1 2 6 A ω 2



L 0

2 m(x)ψ ¯ 1 (x) dx

(8.48)

On equating Tmax from Equation 8.48 to (U)max from Equation 8.46, we get 'L

m(x)ψ ¯ 0 (x)ψ1 (x) dx ω = o' L 2 ¯ 1 (x)} dx o m(x){ψ 2

(8.49)

As in the case of a discrete system, the accuracy of the Rayleigh method can be further improved by using ψ1 (x) to derive a new shape ψ2 (x), which can be used to give a more accurate estimate of the frequency. However, the extra computations involved in the process are hardly justified. Example 8.5 For the frame shown in Example 8.3, obtain improved estimates of the fundamental frequency.

Solution The initial estimate of the displaced shape is given by  ψ T0 = 1

2

3



(a)

The inertia forces Mψ 0 are applied to the frame as shown in Figure E8.5. The resulting displacements are easily computed and are also shown in the figure. We have ψ T1 =

m 4.75 k

9.42

 12.42

Figure E8.5 Frequency of vibration of a three-story shear frame.

(b)

Analysis of single-degree-of-freedom systems 343 Substituting for ψ 0 and ψ 1 in Equation 8.39, we obtain a more accurate estimate of the frequency k ω2 = 0.2246 m ) k ω = 0.474 m A further improvement is obtained in the estimate of frequency, ω, by using Equation 8.41: k ω2 = 0.2237 m ) k ω = 0.473 m

Example 8.6 Find the fundamental frequency of vibration of the cantilever beam shown in Figure E8.6a using the shape function ψ0 (x) = (x/L)2 . Obtain a more accurate estimate of the frequency by the improved Rayleigh method.

Figure E8.6 Vibration of a uniform cantilever beam.

Solution The strain energy of the cantilever is obtained from Equation 8.23: U=

1 2 z 2






L

EI 0

2 L2

2 dx (a)

1 4EI = z2 3 2 L Equation 8.24 gives the kinetic energy T= =

1 2 z˙ 2



L 0

¯ 1 2 mL z˙ 2 5


m ¯

x2 L2

2 dx (b)

344

Dynamics of structures

Assuming simple harmonic motion and that the energy is conserved, we obtain the estimated frequency as ω2 = 20

EI mL3 )

(c)

EI ω = 4.472 mL3  where m is the total mass of the beam. As compared to the exact frequency of 3.5159 EI/mL3 this is 27.2% too high. To obtain a better estimate of the frequency, we apply inertia forces mψ ¯ 0 (x) on the cantilever beam as shown in Figure E8.6b and calculate the deflected shape under the load. Denoting the resulting deflection as y = ψ1 (x), we have from the elementary beam theory EI

d4 y x2 = m ¯ dx4 L2

(d)

Equation d is solved for y with the following two geometric and two natural boundary conditions: At x = 0:

y=0

At x = 0:

dy =0 dx

The two natural boundary conditions require that both the moment and the shear are zero at the free end. The condition that the moment equal to zero at the free end gives x = L:

EI

d2 y =0 dx2

The condition that shear equal to zero at the free end gives x = L:

EI

d3 y =0 dx3

On carrying out the necessary integration, we get the following value for y:
 x6 L2 x2 m ¯ Lx3 + − y = ψ1 (x) = EI 360L2 18 8

(e)

Substitution of ψo and ψ1 in Equation 8.47 gives EI mL3 ) EI ω = 3.53 mL3

ω2 = 12.46

(f)

Analysis of single-degree-of-freedom systems 345 The use of Equation 8.49 provides a further improvement in the value of ω, giving EI mL3 ) EI ω = 3.5164 mL3

ω2 = 12.365

(g)

 This is very close to the exact frequency of 3.5159 EI/mL3 .

8.5

SELECTION OF AN APPROPRIATE VIBRATION SHAPE

Since the accuracy of the frequency estimate obtained by Rayleigh’s method depends on the vibration shape selected, a discussion of the considerations that guide such selection is relevant. We have observed that the generalized coordinate method and the Rayleigh method are essentially the same; the criteria presented in Section 2.7.4 for the selection of a deflected shape therefore apply to the Rayleigh method as well. Briefly, the selected displacement shape should satisfy all of the geometric or essential boundary conditions. Use of vibration shapes that violate one or more of such conditions may lead to estimates of frequency that are inaccurate and unreliable. Additional accuracy is obtained if the selected shapes satisfy some or all of the natural or force boundary conditions. Thus, referring to the vibration of a cantilever beam, the shape ψ = (x/L)2 satisfies the two essential boundary conditions: namely, the deflection and the slope are both zero at the built in end where x = 0. The deflection shape violates the condition that moment should be zero at the free end, but the condition that shear at the free end should be zero is satisfied. For the case of a simply supported beam, the deflection shapes
 w x4 Lx3 L3 x ψ= − + EI 24 12 24 and ψ=

P [3L2 x − 4x3 ] 48EI

both satisfy all four boundary conditions: two on the deflections at supports and two on the moments at those locations. The foregoing discussion suggests that a deflection shape produced by the application of a static load on the system should be an appropriate shape to use in the Rayleigh method, because such a shape will automatically satisfy all boundary conditions. The effectiveness of using a deflected shape produced by application of static load was demonstrated in Example 8.4. The choice has an additional advantage because the calculation of strain energy is simplified, such energy simply being equal to the work done by the static loads in riding through the deflections produced by them. This avoids the need to use the alternative energy expression of Equation 8.23, which involves the second derivative of the assumed shape function. Because the error involved in the use of derivatives of an approximate displacement shape is greater than in the use of the

346

Dynamics of structures

Figure 8.4 Vibrations of a beam with central mass.

shape function itself, the direct method of calculating the strain energy is always more accurate than Equation 8.23. Although, as demonstrated in Example 8.5, any reasonable choice of a static load deflection shape is effective, usually the most appropriate choice is the deflection shape resulting from the application of a gravity load equal to that produced by the mass of the system. As an illustration, consider the simply supported uniform beam with a central mass shown in Figure 8.4. It is required to find the frequency of lateral vibrations of the beam. The deflected shape produced by uniform gravity load mg ¯ and a central load Mg is given by ψ(x) = y1 (x) + y2 (x)

(8.50)

¯ where y1 (x) is the deflection produced by the uniform load mg: y1 (x) =

mg ¯ (x4 − 2Lx3 + L3 x) 24EI

for 0 ≤ x ≤ L

(8.51)

and y2 (x) is the deflection due to the central concentrated load Mg: y2 (x) =

Mg (3L2 x − 4x3 ) 48EI

for 0 ≤ x ≤ L/2

(8.52)

The strain energy stored in the beam is given by Umax

1 = 2

 0

L

1 mgy ¯ 1 (x)dx + 2 2



L/2

mgy ¯ 2 (x)dx

0

1 + Mg{y1 (L/2) + y2 (L/2)} 2   1 M 2 L 3 g 2 2  m 2 5  m  + +1 = 2 48EI 5 M 4 M

(8.53)

where m = mL ¯ is the total mass of the system. The maximum kinetic energy is given by   L/2  1 2 ω 2 m{y ¯ 1 (x) + y2 (x)}2 dx + M{y1 (L/2) + y2 (L/2)}2 2 0  2 3 6 2  62  m 3 113  m 2 243  m  1ω M L g +1 + + = 2 (48EI)2 315 M 112 M 140 M

Tmax =

(8.54)

Analysis of single-degree-of-freedom systems 347

On equating Umax to Tmax , we obtain the following expression for the frequency: ω2 =

(2/5)α 2 + (5/4)α + 1 48EI 3 3 ML (62/315)α + (113/112)α 2 + (243/140)α + 1

(8.55)

where α = m/M. Several points are worth noting. First the gravity constant g drops out from the expression for ω and need not be included in the computations. This is to be expected, because a constant multiplier in the deflection shape does not affect the final results. Second, if the beam is massless, that is, if α = 0, we get ω2 = 48EI/ML3 , which is the exact frequency for a uniform massless beam with a central mass M. The exact result is obtained because y2 (x) is the true vibration shape in this case. Finally, if the central mass is 0, α tends to ∞ and Equation 8.55 gives in the limit EI mL3 ) EI ω = 9.8767 mL3

ω2 = 97.55

(8.56)

which, of course, is the same result as obtained in Example 8.4. It is of interest to note that the deflection shape obtained by applying gravity loading produced by the mass in the system is, in fact, the same as the shape ψ1 obtained by the procedure described in the preceding section when ψ0 is taken to be a constant equal to 1. This is because with ψ0 = 1, the inertia force mψo is proportional to the gravity load mg. When the equivalence between the procedure for obtaining improved shape ψ1 , given ψo = 1, and that for obtaining a deflection shape from mass-produced gravity loads is recognized, it is easy to appreciate that the direction of gravity may have to be adjusted to match the deflected shape with the expected vibration shape. Consider, for example, the beam with overhangs shown in Figure 8.5. The overhangs are expected to move in a direction opposite to that of the span section of the beam. The first estimate of the vibration shape, ψo , should therefore be as indicated in Figure 8.5b. Similarly, the direction of gravity loads should be as shown in Figure 8.5c, where the loads in the overhanging portions are acting in a direction opposite to that of forces within the span. As a final example of the application of mass-produced gravity load, consider a discrete system composed of masses M1 , M2 , . . . , Mn . Let the deflections at the location of the point masses due to gravity loads of magnitudes M1 , M2 , . . . , Mn be equal to y1 , y2 , . . . , yn . The shape function ψ to be used in the Rayleigh method is then given by  ψ T = y1

y2

···

yn



(8.57)

The maximum strain energy is Umax =

1 [M1 y1 + M2 y2 + · · · + Mn yn ] 2

(8.58)

348

Dynamics of structures

Figure 8.5 Vibrations of an overhanging beam: (a) anticipated vibration shape; (b) first estimate of vibration shape, ψ0 ; (c) mass-produced gravity load.

The maximum kinetic energy is given by Tmax =

1 2 ω [M1 y12 + M2 y22 + · · · + Mn yn2 ] 2

(8.59)

Equating Umax to Tmax we obtain the frequency ω: & Mi yi ω2 = & Mi yi2

(8.60)

Note that we have ignored the gravity constant in our computations because it eventually cancels out. Example 8.7 For the frame shown in Example 8.3, calculate the deflected shape resulting from the application of gravity forces equal to the story masses. Use this deflected shape to calculate the frequency of vibration.

Solution The frame represents a discrete system with three concentrated masses and the corresponding three degrees of freedom. Since the vibrations are in the lateral direction, we apply the gravity forces so that they also act in the horizontal direction. The applied forces and the calculations of resulting displacements are indicated in Figure E8.7. The assumed vibration shape then becomes ψT =

m 2.75 k

4.75



5.75

(a)

Analysis of single-degree-of-freedom systems 349

Figure E8.7 Lateral vibrations of a three-story shear frame.

The maximum strain energy is given by m m m 1 m × 5.75 + 2m × 4.75 + 2.5m × 2.75 2 k k k 2 m 1 = × 22.125 2 k

Umax =

(b)

The kinetic energy is Tmax =

1 2 m3 ω 2 × 97.094 2 k

(b)

Equating Umax to Tmax we get k ω2 = 0.2279 m ) k ω = 0.477 m The gravity constant has not been included in the computations because it cancels out.

8.6

SYSTEMS WITH DISTRIBUTED MASS AND STIFFNESS: ANALYSIS OF INTERNAL FORCES

In Section 2.7.4 we discussed a method by which a system with distributed mass and stiffness can be represented by a single-degree-of-freedom system using generalized

350

Dynamics of structures

coordinate and an assumed shape function (Eq. 2.19). The resulting equation for forced damped vibrations works out to m∗ z¨ (t) + c∗ z˙ (t) + k∗ z(t) = p∗ (t)

(8.61)

where m∗ is the generalized mass, c∗ the generalized damping, k∗ the generalized stiffness, and p∗ the generalized force, all determined according to Section 2.7.4. Consider for example the vibrations of the simply supported beam shown in Figure 8.3, and suppose that the equation of motion has been solved to obtain the value of generalized coordinate z(t). The displacement of the beam can be obtained by using Equation 2.19. It is now required to determine the internal forces, namely the bending moments and shears. This can be achieved through a static analysis of the beam for a load that will produce the displaced shape given by z(t)ψ(x). From ordinary beam theory we obtain:   δ2u δ2 δ2 w1 (x, t) = 2 EI(x) 2 = 2 [EI(x)ψ  (x)]z(t) δx δx δx

(8.62)

The equivalent load involves fourth-order differential of the shape function. The shears and moments will, in turn, involve third and second order differentials of the shape function, respectively. The internal forces obtained from the equivalent load are thus less accurate than the displacement because derivatives of the approximate shape function provide poorer approximations than does the shape function itself. A better estimate of the internal forces can be obtained by using the following expression for the equivalent load. w2 (x, t) = ω2 mψ(x)z(t) ¯

(8.63)

Equation 8.63 does not involve derivatives of the shape function and is therefore expected to provide better estimates of the internal forces. The equivalent loads obtained from Equations 8.62 and 8.63 are identical only when the shape function ψ(x) is one of the shapes in which the beam may execute free vibrations. A proof of this is provided in Chapter 15 (Eq. 15.18). A special shape of this kind is known as a mode shape of the beam. When ψ(x) is not one of the mode shapes, loads w1 and w2 are not the same. However the total strain energy stored in the beam as it deflects under the two different loads is the same. An expression for the strain energy stored through deflection u produced by load w1 is given by: U=



1 2

L

w1 (x, t)u(x, t)dx =

0

1 z(t) 2



L

w1 (x, t)ψ(x)dx

(8.64)

0

The strain energy given by Equation 8.64 should be the same as that obtained by considering the work done by the flexural moments (Eq. 8.23). Thus 1 z(t) 2

 0

L

w1 (x, t)ψ(x)dx =

1 2 z (t) 2

 0

L

EI(x){ψ  }2 (x)dx

(8.65)

Analysis of single-degree-of-freedom systems 351

On substituting Equation 8.28 in Equation 8.65 we get 1 z(t) 2



L

0

 L 1 2 2 2 w1 (x, t)ψ(x)dx = z (t)ω m(x)ψ ¯ (x)dx 2 0  L   2 1 = z(t) ¯ ψ(x)dx ω m(x)ψ(x)z(t) 2 0

(8.66)

On comparing the left- and right-hand sides of Equation 8.66 we conclude that an ¯ produces the same strain energy as load equivalent load w2 (x, t) = ω2 m(x)ψ(x)z(t) w1 (x, t). Example 8.8 For the precast concrete T-beam of Example 6.1 calculate the maximum mid-span moment produced due to clapping and stamping by spectators during a sporting event. Use the shape function given in Example 6.1.

Solution The following data has been supplied. E = 3.7 × 1010 N/m2 , k∗ = 701.6 kN/m, ω = 7.38 Hz, ψ(x) =

3

I = 6.327 × 10−3 m4 ,

m∗ = 326.6 kg,

 = 3.0 Hz,

m ¯ = 567.3 kg/m,

ξ = 0.03

p∗ = 0.7137 sin t = p0 sin t kN

β = 0.4065

4

x x x −2 3 + 4 L L L

Using the data supplied we get

z(t) =

p∗ [(1 − β 2 )2 + (2βξ )2 ]−1/2 sin (t − φ) k∗

= 1.218 × 10−3 sin (t − φ) m

(a)

The equivalent load is given by d4 ψ dx4 24 = EIz(t) 4 = 0.365 sin (t − φ) kN/m L

w1 (x, t) = EIz(t)

(b)

The maximum mid-span moment is given by

(M)L/2 =

0.365L2 = 6.247 kN · m 8

(c)

352

Dynamics of structures

As an alternative we may obtain the equivalent load from w2 (x, t) = ω2 mψ(x)z(t) ¯ = 46.352 × 0.5673 × 1.218 × 10−3 ψ(x) sin (t − φ) = 1.484ψ(x) sin (t − φ) kN/m

(d)

The maximum value of the distributed load represented by Equation d is shown in Figure E8.8. The end reaction works out to 1.736 kN and the maximum mid-span moment to 6.447 kN · m. In the present case there is not a large difference between the two estimates of the mid-span moment because the assumed vibration shape is quite close to the first free-vibration mode shape of the beam.

Figure E8.8 Dynamic distributed load on a precast concrete T-beam used to support stadium seating.

8.7

NUMERICAL EVALUATION OF DUHAMEL’S INTEGRAL

As stated earlier, there may be instances when the response history of a dynamic system rather than its frequency is of interest. For a linear system, the response of the system to a specified exciting force is given by the Duhamel’s equation (Eq 8.2). Except when the exciting force is a simple mathematical function, a closed-form solution of Duhamel’s integral is not possible, and a numerical method of evaluation must be used. In this section we present three alternative procedures for the numerical evaluation of Duhamel’s integral and illustrate the computations involved by several examples. For an undamped system the Duhamel’s integral reduces to  t 1 u(t) = p(τ ) sin ω(t − τ )dτ mω 0  t 1 = p(τ )(sin ωt cos ωτ − cos ωt sin ωτ )dτ mω o = A sin ωt − B cos ωt

(8.67)

Analysis of single-degree-of-freedom systems 353

where 1 A= mω 1 B= mω



t

p(τ ) cos ωτ dτ o



(8.68)

t

p(τ ) sin ωτ dτ o

The Duhamel’s integral for a damped system can be reduced as follows:  1 −ξ ωt t ξ ωτ u(t) = e e p(τ )(sin ωd t cos ωd τ − cos ωd t sin ωd τ ) dτ mωd o = Ae−ξ ωt sin ωd t − Be−ξ ωt cos ωd t

(8.69)

where 1 A= mωd 1 B= mωd



t

eξ ωτ p(τ ) cos ωd τ dτ 

o

(8.70)

t

e

ξ ωτ

p(τ ) sin ωd τ dτ

o

It is our objective to carry out a numerical evaluation of the integrals in Equations 8.68 and 8.70. To evaluate A or B we denote the integrand by f (τ ) and plot y = f (τ ) as a function of τ as shown in Figure 8.6a. The required integral is equal to the area enclosed by the graph of f (τ ) and the τ axis between 0 and t. Several alternative procedures are available for the numerical evaluation of the required areas; among them, the following are more commonly used. 1 2 3

Rectangular summation Trapezoidal method Simpson’s method

8.7.1 Rectangular summation To obtain the area enclosed by function f (τ ), we divide the time axis from 0 to t into n small intervals of time, each of duration h. The area under the curve is now approximated by the sum of the areas of N rectangles shown shaded in Figure 8.6b. The ordinate of the first rectangle is y0 and that of the ith is yi−1 . The area of the ith rectangle is hyi−1 . The total area is therefore given by  t y dt = h(y0 + y1 + y2 + · · · + yN−1 ) (8.71) 0

Since the upper boundaries of the rectangles do not match the outline of the curve, the area obtained by the foregoing procedure is in error. In a general case, these errors are self-compensating, as is evident from Figure 8.6b, and provided that h is small enough and a large number of rectangles is used in the summation, reasonable accuracy will be obtained.

354

Dynamics of structures

Figure 8.6 (a) Area under a curve; (b) rectangular summation (c) trapezoidal method; (d) Simpson’s method.

8.7.2 Trapezoidal method As in the case of rectangular summation, the time axis is divided into N equal intervals each of duration h. The portion of the area bounded by the curve, ordinates yi−1 and yi , and the τ axis is now approximated by a trapezium of base h and heights yi−1

Analysis of single-degree-of-freedom systems 355

and yi . The area of the ith trapezium is (h/2)(yi−1 + yi ). The total area is obtained by summing the N trapeziums: 

t

y dτ =

0

h (yo + 2y1 + 2y2 + · · · + 2yN−1 + yN ) 2

(8.72)

It is obvious from a reference to Figure 8.6c that for the same value of h the trapezoidal method will, in general, give a more accurate estimate of the area under the curve y = f (τ ) than rectangular summation would.

8.7.3 Simpson’s method The time axis is again divided into N equal intervals each of duration h. However, N in this case must be an even number. The portion of the curve connecting ordinates y2i−2 , y2i−1 , and y2i is now approximated by a parabola, as shown in Figure 8.6d. It is easily seen that there are N/2 segments of the type shown in the figure and that the area of the ith segment is h (y2i−2 + 4y2i−1 + y2i ) 3 Summation of the areas of N/2 segments gives the total area 

t

y dτ =

0

h (y0 + 4y1 + 2y2 + 4y3 + 2y4 + · · · + 2yN−2 + 4yN−1 + yN ) 3

(8.73)

Example 8.9 The water tower shown in Figure E8.9a is idealized as a single-degree-of-freedom system. It is subjected to the half-sine-wave loading shown in Figure E8.9b. Calculate the displacement history for the first 1.0 s using numerical evaluation of the Duhamel’s integral and h = 0.1 s. Neglect damping and assume that the tower is initially at rest.

Solution The mass m of the tower and the frequency ω are obtained first: 978.8 = 2.533 kip · s2 /in. 386.4 ) ) k 100 = = 6.283 rad/s ω= m 2.533

m=

The displacement response is obtained in terms of Duhamel’s integral as u(t) = A sin ωt − B cos ωt

(a)

where A and B have been defined in Equation 8.68. The response calculations for the first 0.6 s are shown in Tables E8.9a and E8.9b, where results are given for three alternative methods rectangular summation, trapezoidal method, and the Simpson’s method: After 0.6 s, A and B

356

Dynamics of structures

Figure E8.9 Forced vibration of a tower structure: (a) tower; (b) forcing function.

remain unchanged and the response for subsequent intervals of time is given by Equation a, with A and B equal to their values at 0.6 s. Because the forcing function is, in this case, a simple mathematical function, it is possible to obtain a closed-form solution. When the sine-wave function is expressed as P(t) = p0 sin t

(b)

the response for the first 0.6 s is given by

u(t) =

p0 1 (sin t − β sin ωt) k 1 − β2

(c)

where β = /ω. We have π = 5.236 rad/s 0.6 5.236  = = 0.833 β= ω 6.283 p0 = 100 kips =

so that u(t) = 3.267(sin 5.236t − 0.833 sin 6.283t)

(d)

Table E8.9a Numerical evaluation of Duhamel’s equation, undamped system. A¯

B¯

τ (s)

P(τ ) (kips)

cos ωτ

p(τ ) × cos ωτ

Rectangular sum

Trapezoidal method

Simpson’s method

(1)

(2)

(3)

(4)

(5)

(6)

(7)

0.0 0.1 0.2 0.3 0.4 0.5 0.6

0.0 50.0 86.6 100.0 86.6 50.0 0.0

1.000 0.809 0.309 −0.309 −0.809 −1.000 −0.809

0.00 40.45 26.76 −30.90 −70.06 −50.00 0.00

0.00 0.00 40.45 67.21 36.32 −33.74 −83.74

0.0 40.5 107.7 103.5 2.6 −117.5 −167.5

0.0 188.6 21.7 −248.4

p(τ ) × sin ωτ

Rectangular sum

Trapezoidal method

Simpson’s method

(8)

(9)

(10)

(11)

(12)

0.000 0.588 0.951 0.957 0.588 0.000 −0.588

0.00 29.39 82.36 95.10 50.92 0.00 0.00

sin ωτ

0.0 0.0 29.4 111.7 206.8 257.8 257.8

0.0 29.4 141.1 318.6 464.6 515.5 515.5

0.0 199.9 713.6 764.5

0.1 τ = = 6.283 × 10−3 mω 2.533 × 6.283 τ = 3.142 × 10−3 2mω τ = 2.094 × 10−3 3mω

Table E8.9b Numerical evaluation of Duhamel’s equation, undamped system. Rectangular sum

Trapezoidal method

Simpson’s method

t

sin ωt

cos ωt

τ A = A¯ mω

τ B = B¯ mω

A sin ωt −B cos ωt

τ A = A¯ 2mω

τ B = B¯ 2mω

A sin ωt −B cos ωt

0.1 0.2 0.3 0.4 0.5 0.6

0.588 0.951 0.951 0.588 0.000 −0.588

0.809 0.309 −0.309 −0.809 −1.000 −0.809

0.0000 0.2542 0.4223 0.2282 −0.2434 −0.5262

0.0000 0.1847 0.7019 1.2994 1.6199 1.6199

0.0000 0.1847 0.6182 1.1849 1.6199 1.6199

0.1271 0.3382 0.3252 0.0080 −0.3692 −0.5262

0.0923 0.4434 1.0010 1.4597 1.6197 1.6197

0.0000 0.1846 0.6184 1.1856 1.6197 1.6197

τ B = B¯ 3mω

A sin ωt −B cos ωt

0.3949

0.4187

0.2458

0.0454

1.4946

1.2358

−0.5203

1.6013

1.6162

τ A = A¯ 3mω

358

Dynamics of structures Table E8.9c Comparison of numerical solution with theoretical results. t

Duhamel integral A sin ωt−B cos ωt

Closed-form solution, Equation d or e

0.0000 0.1846 0.6184 1.1856 1.6197 1.6197 1.0000 0.0000 −1.0000 −1.6197

0.0333 0.2411 0.6788 1.2290 1.6334 1.6000 0.9880 0.0000 −0.9900 −1.6006

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

At t = 0.6 s, u(t) = 3.267(sin 0.6 × 5.236 − 0.833 sin 0.6 × 6.283) = 1.6 in. u(t) ˙ = 3.267(5.236 cos 0.6 × 5.236 − 0.833 × 6.283 cos 6.283 × 0.6) = −3.272 in./s The exact response subsequent to 0.6 s is given by u(t) = 1.6 cos ω(t − 0.6) −

3.272 sin(t − 0.6) 6.283

(e)

The theoretical response values for the first 1 s obtained from Equations d and e are compared with the results of the numerical evaluation of Duhamel’s integral using the trapezoidal method (Table E8.9c).

Example 8.10 Repeat Example 8.9 with 10% damping in the system. Use the trapezoidal method.

Solution The displacement response is now given by u(t) = A sin ωd t − B cos ωd t where A and B are as in Equation 8.70, and  ωd = ω 1 − ξ 2  = 6.283 1 − 0.12 = 6.2515 rad/s

(a)

Analysis of single-degree-of-freedom systems 359 Table E8.10a Numerical evaluation of Duhamel’s equation, damped system. τ (s)

p(τ ) (kips)

eξ ωτ

cos ωd τ

p(τ )eξ ωτ cos ωd τ

A¯ trapezoidal

sin ωd τ

p(τ )eξ ωτ sin ωd τ

B¯ trapezoidal

0.0 0.1 0.2 0.3 0.4 0.5 0.6

0.0 50.0 86.6 100.0 86.6 50.0 0.0

1.0000 1.0648 1.1339 1.2074 1.2857 1.3691 1.4579

1.0000 0.8109 0.3150 −0.3000 −0.8015 −0.9999 −0.8200

0.000 43.172 30.392 −36.222 −89.240 −68.448 0.000

0.000 43.172 117.276 111.986 −13.476 −171.164 −239.612

0.0000 0.5852 0.9491 0.9540 0.5980 0.0158 −0.5723

0.000 31.156 93.198 115.186 66.823 1.084 0.000

0.000 31.156 155.510 363.894 545.903 613.810 614.894

1 × 0.1 1 τ = = 3.1575 × 10−3 mωd 2 2.533 × 6.2515 × 2

Table E8.10b Numerical evaluation of Duhamel’s equation, damped system. t

A

B

sin ωd t

cos ωd t

e−ξ ωt

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

Equation b

0.8109 0.3150 −0.3000 −0.8015 −0.9999 −0.8200

0.9391 0.8819 0.8282 0.7778 0.7304 0.6850

0.0749 0.3097 0.2792 −0.0198 −0.0063 0.2968

0.0749 0.1363 −0.2853 −1.0737 −1.4144 −1.0912

0.0000 0.1734 0.5645 1.0539 1.4081 1.3880

0.0323 0.2254 0.6204 1.0961 1.4251 1.3772

0.1 0.1362 0.0983 0.5852 0.2 0.3700 0.4907 0.9491 0.3 0.3533 1.1481 0.9540 0.4 −0.0425 1.7224 0.5980 0.5 −0.5401 1.9366 0.0158 0.6 −0.7560 1.9400 −0.5723

e−ξ ωtA sin ωd t e−ξ ωt B cos ωd t

7–8

The response calculations for the first 0.6 s are shown in Tables E8.10a and E8.10b. After 0.6 s, A and B remain unchanged and the response for subsequent intervals of time is given by Equation a with A and B equal to their values at 0.6 s. The exact response to the sine-wave force for the first 0.6 s is given by u(t) =

p0 1 {(1 − β 2 ) sin t − 2βξ cos t} k (1 − β 2 )2 + (2βξ )2  
 ωd2 e−ωξ t  p0 2 t + − 2 t 2βξ cos ω 1 + β sin ω + d d k (1 − β 2 )2 + (2βξ )2 ωd ω2

(b)

where p0 = 100 kips,  = 5.236 rad/s, and β = /ω = 0.833. The response values obtained from Equation b are also shown in Table E8.10b.

8.8

DIRECT INTEGRATION OF THE EQUATIONS OF MOTION

The direct integration of the equations of motion provides the response of the system at discrete intervals of time which are usually equally spaced. Determination of the response involves the computation of three basic parameters: displacement, velocity, and acceleration. The integration algorithms are based on appropriately selected expressions that relate the response parameters at a given interval of time to their

360

Dynamics of structures

values at one or more previous time points. In general, two independent expressions of this nature must be specified. The equation of motion written for the time interval under consideration provides the third expression necessary to determine the three unknown parameters. Direct integration thus involves a marching along the time dimension. It is assumed that at the beginning of the integration, response parameter values have been specified or have been computed at one or more points preceding the time range of interest. These specified or computed values permit the marching scheme to be begun so that response can be computed at as many subsequent points as desired. The accuracy and stability of a scheme depend on the expressions selected for relating the response parameters at a time to their historic value, as well as on the magnitude of the time interval used in the computation. In the following sections we present several methods used in numerical integration of the equation of motion and discuss their relative accuracy as well as the conditions that must be met to guarantee stability of the solution. The following notations are used in the presentation: h = time interval, tn = nh, un = displacement at time nh, u˙ n = velocity at time nh, u¨ n = acceleration at time nh, and pn = the applied forces at time nh. 8.9

INTEGRATION BASED ON PIECE-WISE LINEAR REPRESENTATION OF THE EXCITATION

If the forcing function can reasonably be represented by a series of straight lines, it is possible to develop exact formulas for integration of the equation of motion of a linear system. Consider the time-dependent force p(t) shown in Figure 8.7. Over a sufficiently small interval of time h the variation of force with time can be assumed to be linear as shown by the dashed line in Figure 8.7. With the origin located at time tn , p(τ ) is given by p(τ ) = pn +

pn+1 − pn τ h

(8.74)

An exact solution of the equation of motion exists for the forcing function given by Equation 8.74. Consider first an undamped system. The solution to Equation 8.74

Figure 8.7 Piece-wise linear representation of force p(t).

Analysis of single-degree-of-freedom systems 361

consists of three components: (1) a free-vibration component with initial displacement un and initial velocity u˙ n , (2) forced-vibrations induced by a constant force pn , and (3) forced-vibrations induced by a linearly varying force (pn+1 − pn )τ/h. The freevibration component is given by Equation 5.9 u˙ n sin ωτ ω

u1 (τ ) = un cos ωτ +

(8.75)

The response to a constant force is obtained from Equation 7.14 with ξ = 0. u2 (τ ) =

pn (1 − cos ωτ ) k

(8.76)

The response to linearly varying force is obtained from Equation 7.21. u3 (τ ) =

pn+1 − pn kh


τ−

sin ωτ ω

 (8.77)

The displacement at tn+1 is obtained by adding u1 , u2 , and u3 and substituting h for τ . un+1 = Aun + Bu˙ n + Cpn + Dpn+1

(8.78)

where A = cos ωh sin ωh ω
 1 sin ωh C= − cos ωh k ωh
 1 sin ωh D= 1− k ωh B=

(8.79)

The velocity at tn+1 is obtained by adding the differentials u˙ 1 , u˙ 2 , and u˙ 3 and substituting h for τ . u˙ n+1 = A1 un + B1 u˙ n + C1 pn + D1 pn+1

(8.80)

where A1 = −ω sin ωh B1 = cos ωh
 1 cos ωh 1 C1 = ω sin ωh − + k h h
 1 1 cos ωh D1 = − k h h

(8.81)

362

Dynamics of structures

Recurrence formulas 8.78 and 8.80 permit step-by-step integration provided the displacement and velocity at time t = 0 are given. If the acceleration history is also of interest it can be obtained by satisfying the equation of motion at each time step. Thus u¨ n+1 =

1 (pn+1 − kun+1 ) m

(8.82)

Theoretically, the time interval h may be varied from step to step. However, a uniform time step is usually preferred to save time on the computations. The method is readily extended to damped system. The free-vibration component is now given by Equation 5.37. The response to the constant force pn is obtained from Equation 7.14, and the response to linearly varying force is obtained by solving Equation 7.20. On combining the three components and their derivatives we obtain recurrence formulas that are similar to Equations 8.78 and 8.80 with the coefficients given by the following: ! " ξ −ξ ωh A=e sin ωd h + cos ωd h  1 − ξ2
 1 B = e−ξ ωh sin ωd h ωd ! "  
1 2ξ 1 − 2ξ 2 ξ 2ξ −ξ ωh C= sin ωd h − 1 + +e − cos ωd h k ωh ωd h ωh 1 − ξ2
2   2ξ 2ξ 1 −ξ ωh 2ξ − 1 1− +e sin ωd h + cos ωd h D= k ωh ωd h ωh (8.83) ! " ω A1 = −e−ξ ωh  sin ωd h 1 − ξ2 ! " ξ −ξ ωh B1 = e cos ωd h −  sin ωd h 1 − ξ2 ! "  ω 1 ξ 1 1 −ξ ωh C1 = sin ωd h + cos ωd h +  − +e  k h h 1 − ξ2 h 1 − ξ2 "  ! 1 1 e−ξ ωh ξ D1 = sin ωd h + cos ωd h −  k h h 1 − ξ2 It may be noted that while the integration formulas are exact, approximation is involved in representing the forcing function by a series of straight lines. For a small h the errors involved are expected to be small Example 8.11 Calculate the response of the tower in Example 8.9 to the loading shown there for the first 1.0 s using piece-wise linear representation of the force, as shown in Figure E8.11. Assume that damping in the system is 10% of critical and use h = 0.1 s.

Analysis of single-degree-of-freedom systems 363

Figure E8.11 Piece-wise linear representation of a sinusoidal forcing function.

Solution The following properties have been obtained in Examples 8.9 and 8.10. k = 100 kips/in. m = 2.533 kip · s2 /in. ω = 6.283 rad/s ωd = 6.2515 rad/s u0 = u˙ 0 = u¨ 0 = 0 The coefficients in the recurrence formulas are determined from Equation 8.83. A = 0.81672,

B = 0.08791,

A1 = −3.4706,

B1 = 0.70625,

C = 0.0012073, C1 = 0.016378,

D = 0.00062547, D1 = 0.018328

The displacements and velocities obtained from Equations 8.78 and 8.80, respectively, are shown in Table E8.11. The theoretical response for a sine-wave loading is given by Equation b of Example 8.10. The equation is used to obtain the theoretical response values for the first 0.6 s shown in Table E8.11. For finding the response at time intervals beyond the first 0.6 s, we need the displacement as well as the velocity at t = 0.6 s. The displacement has already been calculated; the velocity is obtained by differentiating Equation b of Example 8.10 and substituting t = 0.6 s. Displacement response beyond 0.6 s is now given by  u(t) = e

−ξ ω(t−0.6)

v0.6 + u0.6 ξ ω sin ωd (t − 0.6) + u0.6 cos ωd (t − 0.6) ωd

 (j)

where u0.6 and v0.6 are the displacement and velocity, respectively, at t = 0.6 s. Theoretical displacement values for t ≥ 0.6 s are calculated from Equation j and are entered in Table E8.11. The difference between the calculated and theoretical values of the response can be attributed to the difference between the actual forcing function and its piece-wise linear representation. To improve the accuracy of computations we need to use a smaller interval of time, so that the piece-wise linear representation more closely matches the actual function.

364

Dynamics of structures

Table E8.11 Piece-wise linear representation of the excitation. un

u˙ n

pn

pn+1

un+1 (Eq. 8.73)

u˙ n+1 (Eq. 8.75)

un (theoretical)

0.0000 0.0313 0.2206 0.6062 1.0713 1.3928 1.3461 0.8510 0.1090 −0.5845 −0.9864

0.0000 0.9164 2.9448 4.5651 4.3453 1.6855 −2.8245 −6.6664 −7.6618 −5.7895 −2.0601

0.0 50.0 86.6 100.0 86.6 50.0 0.0 0.0 0.0 0.0 0.0

50.0 86.6 100.0 86.6 50.0 0.0 0.0 0.0 0.0 0.0

0.0313 0.2206 0.6062 1.0713 1.3928 1.3461 0.8510 0.1090 −0.5845 −0.9864

0.9164 2.9448 4.5651 4.3453 1.6855 −2.8245 −6.6664 −7.6618 −5.7895 −2.0601

0.0000 0.0323 0.2254 0.6204 1.0961 1.4251 1.3772 0.8683 0.1105 −0.5974 −1.0073

Time 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

8.10

DERIVATION OF GENERAL FORMULAS

As mentioned in Section 8.8, time integration methods are based on appropriately selected expressions relating the response parameters at a given interval of time to their values at one or more previous points. A general expression of this type can be written as un+1 =

n

Al ul +

l=n−k

n+1

l=n−k

Bl u˙ l +

n+1

Cl u¨ l + R

(8.84)

l=n−k

where R is a remainder term representing the error in the expression, and Al , Bl , and Cl are constants, some of which may equal zero. Equation 8.84 relates the displacement, velocity, and acceleration at time point n + 1, to their values at the previous points n − k, n − k + 1, . . . , n. The equation has m = 5 + 3k undetermined constants A, B, and C. By a suitable choice for the values of these constants, the equation can be made exact in the special case where u is a polynomial of order m − 1. When the formula is exact , the remainder term R will equal zero. Now, if Equation 8.84 is exact for a polynomial of order (m − 1), it will also be exact when u takes any one of the following values 1, t, t 2 , . . . , t m−1 Therefore, to evaluate the m constants, we can successively set u equal to each of these values in turn, and in each case make a substitution in Equation 8.84. This procedure yields m simultaneous equations involving the set of m coefficients Al , Bl , and Cl . A solution of these equations provides the required values for the coefficients. Now if u = t m is substituted into Equation 8.84, R cannot be expected to equal zero automatically. We will designate the resulting value of R by Em . It will be evident that the larger the number of terms in Equation 8.84, the smaller will be the magnitude of truncation error R. Equations having many terms may, however, encounter spurious roots and instability, characteristics that we shall examine in some detail later. It may

Analysis of single-degree-of-freedom systems 365

not, therefore, be desirable to use all m free constants to reduce the truncation error. An approach having practical usefulness is to employ Equation 8.84 to represent exactly a polynomial of order p − 1, p being an integer smaller than m. Then (m − p) constants become available which can be assigned arbitrary values chosen so as to improve the stability or convergence characteristics of the resulting formula. A formula that is exact for polynomials up to an order m − 1 has a truncation error Em when u = t m . It is of interest to estimate the truncation error term when u is a polynomial of order higher than m, or in the more general case, when u is a function other than a polynomial. Standard textbooks on numerical analysis show that in most cases the truncation error term R is given by R=

Em (m) u (ξ ), m!

(n − k)h ≤ ξ ≤ (n + 1)h

(8.85)

where u(m) is the value of the mth differential of u at t = ξ . Parameter ξ is, in general, not known, and it may not therefore be possible to obtain a precise value for R. Nevertheless, Equation 8.85 can often be used to provide an upper bound on R. An estimate of the truncation error term may be helpful in choosing one formula over the other, although such error is not the only criterion in the choice of a formula. The procedure described in the foregoing paragraph is a very general method of finding formulas. In fact, all formulas of type Equation 8.84 that are used in structural dynamics for time integration can be derived using the procedure described. However, very often such formulas are obtained from physical considerations, such as, for example, an assumed variation in the acceleration u, ¨ or from the well-known finite difference approximations of the differentials. In the following sections, we develop, both from the underlying physical considerations, and by using the technique of this section, a number of more commonly used methods of numerical integration of the equation of motion. In each case we also derive the truncation error term.

8.11

CONSTANT-ACCELERATION METHOD

In deriving the formulas for this method, we make the assumption that over a small interval of time h, the acceleration of the system is constant and is equal to its value at the beginning of the interval, as shown in Figure 8.8. For simplicity and without loss of generality, we shift the origin to the time point n. The constant-acceleration assumption gives u¨ = u¨ n

(8.86)

On integrating Equation 8.86, we get u˙ = u¨ n t + A

(8.87)

where A is an arbitrary constant, whose value is obtained from boundary condition u˙ = u˙ n

for t = 0

(8.88)

366

Dynamics of structures

Figure 8.8 Constant-acceleration method.

Substitution of Equation 8.88 in Equation 8.87 gives A = u˙ n . Now setting t = h in Equation 8.87 and recognizing that for this value of t, u˙ = u˙ n+1 , we have u˙ n+1 = u˙ n + hu¨ n

(8.89)

Integrating Equation 8.87 and using the boundary condition u = un at t = 0, we get u = un + u˙ n t +

u¨ n t 2 2

(8.90)

For t = h, u = un+1 , so that Equation 8.90 gives un+1 = un + hu˙ n +

h2 u¨ n 2

(8.91)

We next derive expressions similar to those of Equations 8.89 and 8.91 by the method of Section 8.10. The velocity expression can be written as u˙ n+1 = a1 u˙ n + a2 u¨ n

(8.92)

where a1 and a2 are constants to be determined. Since we have two free constants, we can expect to make Equation 8.92 exact for u = 1 and u = t. For u = 1, u˙ = 0, and u¨ = 0, and Equation 8.92 is automatically satisfied because we get zeros on both sides. For u = t, u˙ = 1, and u¨ = 0. Substitution in Equation 8.92 gives 1 = a1

(8.93)

We are still left with one free constant and can therefore make the formula exact for u = t 2 as well. In this case we have u˙ = 2t and u¨ = 2, and substitution into Equation 8.92 gives 2h = 0 + 2a2

(8.94)

Analysis of single-degree-of-freedom systems 367

Equations 8.92, 8.93, and 8.94 lead to the expression u˙ n+1 = u˙ n + hu¨ n

(8.95)

which is the same as Equation 8.89. Now if we substitute u = t 3 , we cannot expect Equation 8.95 to be satisfied. In fact, a remainder term is obtained in this case. We denote this term by E3 , so that 3h2 = E3

(8.96)

Finally, using Equation 8.85, we write Equation 8.95 with the error term included E3 (3) u (ξ ) 3! h2 = u˙ n + hu¨ n + u(3) (ξ ) 2

u˙ n+1 = u˙ n + hu¨ n +

(8.97)

Next we write the displacement expression as un+1 = b1 un + b2 u˙ n + b3 u¨ n + R

(8.98)

On making Equation 8.98 exact for u = 1, t, t 2 , we get u=1

1 = b1

u=t

h = b2

u=t

2

(8.99)

h = 2b3 2

Substitution of Equation 8.99 into Equation 8.98 gives un+1 = un + hu˙ n +

h2 u¨ n + R 2

(8.100)

To obtain the error term R, we try u = t 3 . Substitution into Equation 8.98 gives h3 = E3

(8.101)

From Equation 8.85, R=

E3 (3) h3 (3) u (ξ ) = u (ξ ) 3! 6

(8.102)

Equations 8.89 and 8.91 provide two of the three relations required for time integration. The third relationship is obtained by writing the equation of motion at time point n + 1: mu¨ n+1 + cu˙ n+1 + kun+1 = pn+1

(8.103)

368

Dynamics of structures

Together, Equations 8.89, 8.91, and 8.103 allow us to obtain the parameters un+1 , u˙ n+1 , and u¨ n+1 in terms of un , u˙ n and u¨ n . Equation 8.89 gives velocity u˙ n+1 , while Equation 8.91 gives displacement un+1 . Substituting for un+1 and u˙ n+1 in Equation 8.103, we get the acceleration u¨ n+1 : u¨ n+1 =


   kh2 1 pn+1 − kun − (c + kh)u˙ n − ch + u¨ n m 2

(8.104)

To begin the time integration, we need to know the values of u0 , u˙ 0 , and u¨ 0 , the displacement, velocity, and acceleration, respectively, at time t = 0. Two of them must be specified; the third is obtained by using the equation of motion at t = 0.

8.12

NEWMARK’S β METHOD

In 1959, Newmark devised a series of numerical integration formulas which are collectively known as Newmark’s β methods. We derive Newmark’s formulas by the general method of Section 8.10. The velocity expression is of the form u˙ n+1 = a1 u˙ n + a2 u¨ n + a3 u¨ n+1

(8.105)

Equation 8.105 is automatically satisfied with u = 1, so that we still have three free constants, a1 , a2 , and a3 . We use two of them to make the formula exact for u = t and t 2 , leaving one of the constants slack. Substitution into Equation 8.105 gives u=t

1 = a1

u = t2

2h = 2a2 + 2a3

(8.106)

We now have two equations and three unknowns. We assign an arbitrary value to one of the unknowns and determine the other two from Equation 8.106. Let us select a3 = γ h, where γ is an arbitrary constant. We then get a1 = 1 a2 = h(1 − γ )

(8.107)

Equation 8.105 now takes the form u˙ n+1 = u˙ n + h(1 − γ )u¨ n + hγ u¨ n+1 + R

(8.108)

The error term is obtained by substituting u = t 3 in Equation 8.108, so that 3h2 = 6h2 γ + E3 E3 = h2 (3 − 6γ )

(8.109a)

Analysis of single-degree-of-freedom systems 369

Using Equations 8.85 and 8.109a, we get E3 (3) u (ξ ) 3! 
1 − γ u(3) (ξ ) = h2 2

R=

(8.109b)

Constant γ is usually selected to be 1/2. With this value for γ , Equation 8.108 becomes u˙ n+1 = u˙ n +

h h u¨ n + u¨ n+1 + R 2 2

(8.110)

Further, the error term given by Equation 8.109b becomes zero. This implies that the formula is exact for polynomials of order up to 3. The new error term is obtained by substituting u = t 4 and can be shown to be R=−

h3 (4) u (ξ ) 12

(8.111)

The displacement expression in Newmark’s β method is un+1 = b1 un + b2 u˙ n + b3 u¨ n + b4 u¨ n+1

(8.112)

Equation 8.112 has four free constants. We use three of them to make the formula exact for u = 1, t, and t 2 . Substituting these values of u, in turn, in Equation 8.112, we obtain u=1

1 = b1

u=t

h = b2

u=t

2

(8.113)

h = 2b3 + 2b4 2

Equation 8.113 provides three relationships among four unknowns. We assign an arbitrary value βh2 to b4 so that b3 = h2 ( 12 − β). Equation 8.112 now reduces to
un+1 = un + hu˙ n + h

2

 1 − β u¨ n + h2 β u¨ n+1 + R 2

(8.114)

The error term R is obtained by substituting u = t 3 in Equation 8.114, which gives h3 = 6βh3 + E3 E3 = h3 (1 − 6β)

(8.115a)

370

Dynamics of structures

and E3 (3) u (ξ ) 3!
 1 = h3 − β u(3) (ξ ) 6

R=

(8.115b)

By assigning different values to γ and β, a series of integration formulas can be obtained. As an example, when γ = 0 and β = 0, Newmark’s β method reduces to the constant-acceleration method, which we have already discussed. Other more commonly used versions of the Newmark’s β method are: 1 2

The average acceleration method, γ = 12 , β = 14 The linear acceleration method, γ = 12 , β = 16

A description of these methods is presented in the following paragraphs.

8.12.1 Average acceleration method When γ is set equal to

1 2

and β = 14 , Equations 8.108 and 8.114 reduce to

h h3 (4) (u¨ n + u¨ n+1 ) − u (ξ ) 2 12 h2 un+1 = un + hu˙ n + (u¨ n + u¨ n+1 ) + 0.152h3 u(3) (ξm ) 4 u˙ n+1 = u˙ n +

(8.116) (8.117)

The error term in Equation 8.116 has been obtained from Equation 8.111. For β = 14 , the error expression of Equation 8.115b is not applicable. The error term in Equation 8.117, obtained by a somewhat more complicated procedure,1 represents an upper bound. The coordinate value ξm lies between 0 and h and is selected to give the maximum value of differential u(3) . Equations 8.116 and 8.117 can also be derived by assuming that, as indicated in Figure 8.9, the acceleration of the system remains constant over the small interval h and its value is equal to the average of the values of accelerations at the beginning and end of the interval: u¨ =

1 (u¨ n + u¨ n+1 ) 2

(8.118)

For simplicity, we shift the origin on the time axis to point tn . Then by integrating Equation 8.118 and applying the boundary conditions u˙ = u˙ n at t = 0 and u˙ = u˙ n+1 at t = h, we get u˙ = u˙ n +

t (u¨ n + u¨ n+1 ) 2

(8.119)

1 R. W. Hamming, Numerical Methods for Scientists and Engineers (New York: McGraw Hill Book Co. Inc., 1962).

Analysis of single-degree-of-freedom systems 371

Figure 8.9 Average acceleration method.

and u˙ n+1 = u˙ n +

h (u¨ n + u¨ n+1 ) 2

(8.120)

Integration of Equation 8.119 with the boundary conditions u = un at t = 0 and u = un+1 at t = h yields un+1 = un + u˙ n h +

h2 (u¨ n + u¨ n+1 ) 4

(8.121)

Equations 8.120 and 8.121 are, of course, the same as Equations 8.116 and 8.117, respectively. Equations 8.120 and 8.121 combined with Equation 8.103 enable us to solve for un+1 , u˙ n+1 , and u¨ n+1 in terms of un , u˙ n , and u¨ n . From Equation 8.121 we obtain u¨ n+1 =

4 (un+1 − un − hu˙ n ) − u¨ n h2

(8.122)

Substitution of Equation 8.122 into Equation 8.120 gives u˙ n+1 = −u˙ n +

2 (un+1 − un ) h

(8.123)

Finally on substituting for u¨ n+1 and u˙ n+1 from Equations 8.122 and 8.123 in Equation 8.103 and collecting terms, we get 
 

2 4 4 4m 2c (8.124) + k un+1 = pn+1 + m un + u˙ n + un + u˙ n + u¨ n + c h2 h h2 h h Equation 8.124 can be solved for un+1 . Substitution into Equations 8.122 and 8.123 then gives u¨ n+1 and u˙ n+1 , respectively. To begin the integration, u0 , u˙ 0 , and u¨ 0 must be known. Two of these, usually the initial displacement and the initial velocity, will be given; the third can be determined by using the equation of motion written at t = 0.

372

Dynamics of structures

Figure 8.10 Linear acceleration method.

8.12.2 Linear acceleration method With γ = 12 and β = 16 , Equations 8.108 and 8.114 reduce to h h3 (4) (u¨ n + u¨ n+1 ) − u (ξ ) 2 12 h2 h2 h4 (4) u (ξ ) un+1 = un + hu˙ n + u¨ n + u¨ n+1 − 3 6 24

u˙ n+1 = u˙ n +

(8.125) (8.126)

The error term in Equation 8.125 is obtained from Equation 8.111. However, when Equation 8.115 is used to obtain the error term for the displacement relation, a zero value is obtained. This is because Equation 8.126 is exact for a polynomial of order up to 3. The error term can, however, be obtained by standard procedure, first substituting u = t 4 in Equation 8.126 to obtain E4 and then by using Equation 8.85. Equations 8.125 and 8.126 can also be derived by assuming that, as indicated in Figure 8.10, the acceleration of the system varies linearly over a small interval of time h. With the origin centered on the time axis at tn , we have u¨ = u¨ n +

u¨ n+1 − u¨ n t h

(8.127)

Integration of Equation 8.127 and substitution of the boundary conditions at t = 0 and t = h gives u˙ = u˙ n + u¨ n t + (u¨ n+1 − u¨ n )

t2 2h

(8.128)

and u˙ n+1 = u˙ n +

h (u¨ n + u¨ n+1 ) 2

(8.129)

Analysis of single-degree-of-freedom systems 373

On integrating Equation 8.128 and substituting the boundary conditions, we get un+1 = un + hu˙ n +

h2 h2 u¨ n + u¨ n+1 3 6

(8.130)

Together, Equations 8.129 and 8.130 and the equation of motion written at tn+1 (Eq. 8.103) enable us to solve for un+1 , u˙ n+1 , and u¨ n+1 in terms of un , u˙ n , and u¨ n . Thus Equation 8.130 gives u¨ n+1

6 = 2 h




h2 un+1 − un − hu˙ n − u¨ n 3

 (8.131)

Substitution of Equation 8.131 in Equation 8.129 yields u˙ n+1 =

3 h (un+1 − un ) − 2u˙ n − u¨ n h 2

(8.132)

Finally, on substituting Equations 8.131 and 8.132 in Equation 8.103 and collecting terms, we get




 6m 3c 3 6 6 h + un + u˙ n + 2u¨ n + c + k un+1 = pn+1 + m un + 2u˙ n + u¨ n h2 h h2 h h 2 (8.133)

Equation 8.133 is solved for un+1 . Equations 8.131 and 8.132 then give the acceleration and velocity, respectively, at time tn+1 . Again, to start the time integration, we need to know u0 , u˙ 0 , and u¨ 0 . Two of these three parameters will be specified and the third can be obtained from the equation of motion written at t = 0. Example 8.12 Calculate the response of the tower in Example 8.9 to the loading shown there for the first 1.0 s using a numerical integration technique. Assume that damping in the system is 10% of critical and use h = 0.1 s.

Solution We will obtain the response using (a) the constant-acceleration method, (b) the average acceleration method, and (c) the linear acceleration method. For each case, k = 100 kips/in. m = 2.533 kip · s2 /in. ω = 6.283 rad/s c = 2ξ ωm = 3.183 kip · s/in. u0 = u˙ 0 = u¨ 0 = 0 (a) Constant-acceleration method. Using Equations 8.91, 8.89, and 8.104, we get un+1 = un + 0.1u˙ n + 0.005u¨ n

(a)

374

Dynamics of structures

Table E8.12a Constant-acceleration method Time 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

un

u˙ n

u¨ n

pn+1

un+1 (Eq. a)

u˙ n+1 (Eq. b)

u¨ n+1 (Eq. c)

un (theoretical)

0.0000 0.0000 0.0987 0.4351 0.9923 1.5660 1.8320 1.4930 0.5400 −0.6950 −1.7430

0.000 0.000 1.974 4.755 6.387 5.086 0.239 −7.024 −12.036 −12.655 −8.322

0.00 19.74 27.81 16.32 −13.01 −48.47 −72.63 −50.11 −6.19 43.33 79.29

50.0 86.6 100.0 86.6 50.0 0.0 0.0 0.0 0.0 0.0

0.0000 0.0987 0.4351 0.9922 1.5659 1.8322 1.4927 0.5401 −0.6949 −1.7430

0.000 1.974 4.755 6.387 5.086 0.239 −7.024 −12.036 −12.655 −8.322

19.74 27.81 16.32 −13.01 −48.47 −72.63 −50.11 −6.19 43.33 79.29

0.0000 0.0323 0.2254 0.6204 1.0961 1.4251 1.3772 0.8683 0.1105 −0.5974 −1.0073

u˙ n+1 = u˙ n + 0.1u¨ n u¨ n+1 =

1 (pn+1 − 100un − 13.183u˙ n − 0.8183u¨ n ) 2.533

(b) (c)

Step-by-step time integration is now carried out using Equations a, b, and c. Response calculations for the first 1.0 s are shown in Table E8.12a. (b) Average acceleration method. Substituting the values of m, c, k, and h in Equation 8.124, we get 1176.9un+1 = pn+1 + 1076.9un + 104.5u˙ n + 2.533u¨ n

(d)

Equations 8.123 and 8.122 give u˙ n+1 = 20(un+1 − un ) − u˙ n

(e)

u¨ n+1 = 400(un+1 − un ) − 40u˙ n − u¨ n

(f)

Time integration is carried out using Equations d, e, and f. Response calculations for the first 1.0 s are shown in Table E8.12b. (c) Linear acceleration method. Substitution for m, c, k, and h in Equation 8.133 gives 1715.3un+1 = pn+1 + 1615.3un + 158.35u˙ n + 5.225u¨ n

(g)

Equations 8.132 and 8.131 give u˙ n+1 = 30(un+1 − un ) − 2u˙ n − 0.05u¨ n

(h)

u¨ n+1 = 600(un+1 − un ) − 60u˙ n − 2u¨ n

(i)

Time integration is carried out using Equations g, h, and i. Response calculations for the first 1.0 s are shown in Table E8.12c. The theoretical response for a sine-wave loading is given by Equation b of Example 8.10. The equation is used to obtain the theoretical response values for the first 0.6 s shown in Tables E8.12a, E8.12b, and E8.12c. Theoretical displacement values for t ≥ 0.6 s are calculated

Analysis of single-degree-of-freedom systems 375 Table E8.12b Average acceleration method. Time 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

un

u˙ n

u¨ n

pn+1

un+1 (Eq. d)

u˙ n+1 (Eq. e)

u¨ n+1 (Eq. f)

un (theoretical)

0.0000 0.0425 0.2245 0.5851 1.0248 1.3432 1.3291 0.9073 0.2227 −0.4633 −0.9175

0.0000 0.8497 2.7902 4.4218 4.3736 1.9943 −2.2767 −6.1607 −7.5297 −6.1909 −2.8928

0.000 16.994 21.816 10.815 −11.779 −35.806 −49.613 −28.066 0.686 26.089 39.871

50.0 86.6 100.0 86.6 50.0 0.0 0.0 0.0 0.0 0.0

0.0425 0.2245 0.5851 1.0248 1.3433 1.3291 0.9073 0.2227 −0.4633 −0.9175

0.8497 2.7902 4.4218 4.3736 1.9943 −2.2767 −6.1607 −7.5297 −6.1909 −2.8928

16.994 21.816 10.815 −11.779 −35.806 −49.613 −28.066 0.686 26.089 39.871

0.0000 0.0323 0.2254 0.6204 1.0961 1.4251 1.3772 0.8683 0.1105 −0.5974 −1.0073

Table E8.12c Linear acceleration method. Time 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

un

u˙ n

u¨ n

pn+1

un+1 (Eq. g)

u˙ n+1 (Eq. h)

u¨ n+1 (Eq. k)

un (theoretical)

0.0000 0.0291 0.2119 0.5896 1.0532 1.3862 1.3644 0.8969 0.1678 −0.5389 −0.9785

0.0000 0.8745 2.8603 4.4991 4.3819 1.8707 −2.5299 −6.4303 −7.6476 −6.0547 −2.5210

0.000 17.490 22.227 10.549 −12.893 −37.331 −50.680 −27.328 2.981 28.875 41.789

50.0 86.6 100.0 86.6 50.0 0.0 0.0 0.0 0.0 0.0

0.0291 0.2119 0.5896 1.0532 1.3862 1.3644 0.8969 0.1678 −0.5389 −0.9785

0.8745 2.8603 4.4991 4.3819 1.8707 −2.5299 −6.4303 −7.6476 −6.0547 −2.5210

17.490 22.227 10.549 −12.893 −37.331 −50.680 −27.328 2.981 28.875 41.789

0.0000 0.0323 0.2254 0.6204 1.0961 1.4251 1.3772 0.8683 0.1105 −0.5974 −1.0073

from Equation j of Example 8.11 and are entered in Tables E8.12a, E8.12b, and E8.12c. An examination of the response values in these tables shows that the average acceleration and the linear acceleration methods give reasonable results. The time step h of 0.1 s, which is one-tenth the natural period of the system and one-twelfth the period of exciting force is probably the maximum one can use to ensure a reasonable accuracy; a smaller step size must be used if better accuracy is desired. The results obtained from the constant-acceleration method are not satisfactory. A considerably smaller step size must be used in this case to obtain acceptable accuracy. In fact, the method is not very effective, for several reasons that will be discussed later.

8.13 WILSON-θ METHOD This method proposed by E. L. Wilson is similar to the linear acceleration method. It is based on the assumption that the acceleration varies linearly over an extended interval θ h as shown in Figure 8.11. Parameter θ, which is always greater than 1, is selected to give the desired characteristics of accuracy and stability.

376

Dynamics of structures

Figure 8.11 Wilson-θ method.

The basic relationships used in the method are similar to Equations 8.131, 8.132, and 8.133 with n + 1 replaced by n + θ, and h replaced by θ h.   6 (θh)2 u (8.134) − u − θ h u ˙ − u ¨ u¨ n+θ = n+θ n n n (θh)2 3 u˙ n+θ =

3 θh (un+θ − un ) − 2u˙ n − u¨ n θh 2

(8.135)

     3 6un 6 θh 6m 3c + +k un+θ = pn+θ +m + u˙ n +2u¨ n +c un +2u˙ n + u¨ n (θh)2 θ h (θ h)2 θh θh 2



(8.136) Since the acceleration is assumed to vary linearly from time nh to (n + θ)h, the exciting force is also assumed to vary linearly over the same interval of time and pn+θ is therefore obtained by projecting the exciting force value to time (n + θ)h. Thus pn+θ = pn +

pn+1 − pn θh h

= pn (1 − θ ) + pn+1 θ

(8.137)

Equation 8.136 is used to calculate un+θ . This value is then substituted in Equation 8.134 to obtain u¨ n+θ . The acceleration at normal time increment h is now given by u¨ n+1 = u¨ n +

u¨ n+θ − u¨ n θ

(8.138)

This value of u¨ n+1 is substituted in Equations 8.129 and 8.130 to obtain the velocity and displacement, respectively, at normal increment of time. Example 8.13 Solve Example 8.12 by the Wilson θ method using θ = 1.5 and h = 0.1 s.

Analysis of single-degree-of-freedom systems 377 Table E8.13 Wilson-θ method. Time

u˙ n

un

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.0000 0.0265 0.1932 0.5419 0.9827 1.3265 1.3668 1.0010 0.3634 −0.3174 −0.8252

u¨ n

pn+θ

u¨ n+θ (Eq. b)

un+θ (Eq. a)

0.0000 0.00 75.0 0.0894 0.7946 15.89 104.9 0.3508 2.6181 20.58 106.7 0.7634 4.1966 10.99 79.9 1.1806 4.2791 −9.34 31.7 1.3934 2.2254 −31.73 −25.0 1.2237 −1.6564 −45.91 0.0 0.7011 −5.3658 −28.28 0.0 0.0143 −6.9832 −4.07 0.0 −0.6018 −6.2529 18.67 0.0 −0.9640 −3.6641 33.11

u¨ n+1 (Eq. c)

u˙ n+1 (Eq. d)

un+1 (Eq. e)

23.837 15.89 0.7946 0.0265 22.923 20.58 2.6188 0.1932 6.197 10.99 4.1966 0.5419 −19.507 −9.34 4.2791 0.9827 −42.927 −31.73 2.2254 1.3265 −52.993 −45.91 −1.6564 1.3668 −19.467 −28.28 −5.3658 1.0010 8.039 −4.07 −6.9832 0.3634 30.041 18.67 −6.2529 −0.3174 40.323 33.11 −3.6641 −0.8252

un (theoretical) 0.0000 0.0323 0.2254 0.6204 1.0961 1.4251 1.3772 0.8683 0.1105 −0.5974 −1.0073

Solution Substitution of m, c, k, h, and θ in Equation 8.136 gives 839.13un+θ = pn+θ + 739.13un + 107.69u˙ n + 5.305u¨ n

(a)

where pn+θ is obtained from Equation 8.137. Equation 8.134 now leads to u¨ n+θ = 266.7(un+θ − un ) − 40u˙ n − 2u¨ n

(b)

From Equation 8.138 we get u¨ n+1 = 0.6667u¨ n+θ + 0.3333u¨ n

(c)

Finally, substitution of Equation c in Equations 8.129 and 8.130 gives u˙ n+1 = u˙ n + 0.05(u¨ n + u¨ n+1 )

(d)

un+1 = un + 0.1u˙ n + 0.00333u¨ n + 0.00167u¨ n+1

(e)

Step-by-step time integration is carried out using Equations a through e. Response calculations for the first 1 s are shown in Table E8.13.

8.14

METHODS BASED ON DIFFERENCE EXPRESSIONS

8.14.1 Central difference method The method uses standard central difference expressions to relate the time derivatives of displacement, that is, velocity and acceleration, to the displacement values at selected intervals of time. The velocity expressions is given by u˙ n =

1 (un+1 − un−1 ) + R 2h

(8.139)

378

Dynamics of structures

A graphical representation of Equation 8.139 is shown in Figure 8.12a. Substitution of u = 1, t, t 2 , in turn, in Equation 8.139 shows that the equation is exact for these values of u, so that the remainder term is zero. For u = t 3 , we get R = E3 = −h2 . In a general case, therefore, R=

h2 E3 (3) u (ξ ) = − u(3) (ξ ) 3! 6

(8.140)

The central difference expression used for acceleration is u¨ n =

1 (un+1 − 2un + un−1 ) + R h2

Figure 8.12 Central difference method.

(8.141)

Analysis of single-degree-of-freedom systems 379

A graphical representation of Equation 8.141 is shown in Figure 8.12b. By substitution it can be shown that Equation 8.141 is exact for u = 1, t, t 2 , t 3 . For u = t 4 , we get R = E4 = −2h2 . In a general case, therefore, R=

h2 E4 (4) u (ξ ) = − u(4) (ξ ) 4! 24

(8.142)

The equation of dynamic equilibrium is formed at time point n, mu¨ n + cu˙ n + kun = pn

(8.143)

Substituting Equations 8.139 and 8.141 in Equation 8.143 and solving for un+1 , we get 
 c c  2m m u + = p + −k + + un−1 u − n+1 n n h2 2h h2 2h h2

m

(8.144)

Equations 8.139 and 8.141 are now used to find u˙ n and u¨ n . It will be evident from Equation 8.144 that to start the time integration, we need the values of both u0 and u−1 . When Equations 8.139 and 8.141 are written at t = 0 and solved simultaneously, the following expression is obtained for u−1 : u−1 = u0 +

h2 u¨ 0 − hu˙ 0 2

(8.145)

In a well-posed problem, two of the three parameters u0 , u˙ 0 , and u¨ 0 will be specified; the third can be obtained from the equation of motion (Eq. 8.143) written at t = 0. It will be observed that in all the methods discussed so far, except in the central difference method, one of the three expressions used in the solution procedure was the equation of motion written at time point t = (n + 1)h. In contrast, the central difference method uses the equation of motion at t = nh. Methods that use the equation of motion at n + 1 are called implicit methods, while those that use the equation at n are called explicit methods. As we shall see later, in the case of a multi-degree-of-freedom system, use of an implicit method may require much more computation than that needed for an explicit method. On the other hand, while explicit methods are only conditionally stable, some of the implicit methods are unconditionally stable. Conditional stability may be a serious restriction in the analysis of some multi-degree-of-freedom systems. Stability of integration methods is discussed in Section 8.16. Example 8.14 Solve Example 8.12 by the central difference method with h = 0.1 s.

Solution On substituting for m, c, k, and h, Equation 8.144 becomes 269.21un+1 = pn + 406.6un − 237.39un−1

(a)

380

Dynamics of structures

Table E8.14 Central difference method. Time 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

pn

un−1

un

0.0 50.0 86.6 100.0 86.6 50.0 0.0 0.0 0.0 0.0 0.0

0.0000 0.0000 0.0000 0.1857 0.6022 1.1172 1.4780 1.4329 0.8609 0.0366 −0.7038

0.0000 0.0000 0.1857 0.6022 1.1172 1.4780 1.4329 0.8609 0.0366 −0.7038 −1.0953

un+1 (Eq. a)

u˙ n (Eq. b)

u¨ n (Eq. c)

un (theoretical)

0.0000 0.1857 0.6022 1.1172 1.4780 1.4329 0.8609 0.0366 −0.7038 −1.0953

0.0000 0.9286 3.0110 4.6574 4.3792 1.5785 −3.0859 −6.9813 −7.8231 −5.6593

0.000 18.573 23.074 9.854 −15.418 −40.594 −52.693 −25.216 8.381 34.893

0.0000 0.0323 0.2254 0.6204 1.0961 1.4251 1.3772 0.8683 0.1105 −0.5974 −1.0073

Equations 8.139 and 8.141 then give u˙ n = 5(un+1 − un−1 )

(b)

u¨ n = 100(un+1 − 2un + un−1 )

(c)

To start the integration, Equation 8.145 is used to obtain u−1 . In this case because u0 , u˙ 0 , and u¨ 0 are all zero, u−1 is also zero. Equations a, b, and c are now used to carry out step-by-step integration. Response calculations are shown in Table E8.14 for the first 1 s. If the velocity and acceleration are not of interest, the corresponding columns may be omitted from the table.

8.14.2 Houbolt’s method Houbolt’s method provides another example of a method that uses difference operators to represent the time derivatives of displacement. The method employs double backward difference operators to obtain expressions for velocity and acceleration. The velocity expression is of the form u˙ n+1 = a1 un−2 + a2 un−1 + a3 un + a4 un+1

(8.146)

Equation 8.146 has four free constants and can be made exact for u = 1, t, t 2 , and t 3 . Substitution of these values in turn leads to the following relationships: 0 = a1 + a2 + a3 + a4 1 = −2ha1 − ha2 + ha4 2h = 4h2 a1 + h2 a2 + h2 a4

(8.147)

3h2 = −8h3 a1 − h3 a2 + h3 a4 Simultaneous solution of Equation 8.147 gives a1 = −1/(3h), a2 = 3/(2h), a3 = −3/h, and a4 = 11/(6h). Equation 8.146 therefore becomes u˙ n+1 =

1 (−2un−2 + 9un−1 − 18un + 11un+1 ) 6h

(8.148)

Analysis of single-degree-of-freedom systems 381

The acceleration expression is of the form u¨ n+1 = b1 un−2 + b2 un−1 + b3 un + b4 un+1

(8.149)

Equation 8.149 also has four free constants and can be made exact for u = 1, t, t 2 , and t 3 . Substitution in turn into Equation 8.149 leads to 0 = b1 + b2 + b3 + b4 0 = −2hb1 − hb2 + hb4 2 = 4h2 b1 + h2 b2 + h2 b4

(8.150)

6h = −8h3 b1 − h3 b2 + h3 b4 Solution of the set of Equations 8.150 gives b1 = −1/h2 , b2 = 4/h2 , b3 = −5/h2 , and b4 = 2/h2 . Equation 8.149 therefore becomes u¨ n+1 =

1 ( − un−2 + 4un−1 − 5un + 2un+1 ) h2

(8.151)

The third expression used in Houbolt’s method is the equation of motion written at t = (n + 1)h. Simultaneous solution of Equations 8.148 and 8.151 and the equation of motion gives

 2m 11c 5un 4un−1 un−2 + = p + m − + + k u n+1 n+1 h2 6h h2 h2 h2
 3un 3un−1 un−2 +c − + (8.152) h 2h 3h It will be observed from Equation 8.152 that u0 , u1 , and u2 must be known or determined before time integration can be started. Displacement u0 will be specified, while u1 and u2 are calculated by one of the other methods described earlier, such as, for example, the average acceleration method or the linear acceleration method. Houbolt’s method belongs to the class of implicit methods and, as shown later, is unconditionally stable.

8.15

ERRORS INVOLVED IN NUMERICAL INTEGRATION

The errors involved in the numerical integration of differential equations, such as for example an equation of motion, can be classified into three types: 1 2

3

Round-off errors introduced by repeated computations using a small step size. Truncation errors involved in representing un+1 or u˙ n+1 by a finite number of terms in the Taylor series expansion. This is the error term represented by R in the preceding sections. Propagated error introduced by replacing the differential equation by a finite difference equivalent.

382

Dynamics of structures

We will not consider round-off errors in detail here. They are random in nature and therefore must be treated by statistical methods. Error bounds obtained by statistical means often grossly overestimate the real errors. In computer calculations it is often possible to reduce the round-off errors by using a higher precision in the computations. Truncation errors are accumulated locally at each step. Unless the integration method being used is unstable, truncation error is a good indication of the accuracy of the numerical solution. Truncation error terms therefore provide a useful criterion for assessing the relative accuracy of the various methods of numerical integration. An estimate of the truncation error in a formula can be obtained by considering the equation of undamped free vibration: u¨ + ω2 u = 0

(8.153)

Equation 8.153 has a solution of the type u = c1 sin ωt + c2 cos ωt

(8.154)

in which c1 and c2 are the constants of integration. In the integration formulas presented in this chapter, the truncation term is, in general, of the form Chp u(p) (ξ )

(8.155)

where C is a constant and p is the degree of the formula. From Equation 8.154 we note that the pth derivative of u contains a factor ωp ; the magnitude of the truncation error will therefore depend on the value of (ωh)p . It is evident that to keep the truncation error low, the time step size h should be chosen so that ωh is less than 1. Propagated errors affect the stability of the computation and are discussed in detail in the next section.

8.16

STABILITY OF THE INTEGRATION METHOD

As stated in Section 8.15, errors are introduced into the numerical solution due to truncation. It is important to know the effect of the error introduced at one step on the computations at the next step. If the error has a tendency to grow, the solution soon becomes unbounded and meaningless. In such a situation the computational method is said to be unstable. In a general discussion of the stability of integration methods, it is convenient to use certain matrix notations. We review here briefly those terms in matrix algebra that we shall find useful. Let A be a square matrix of order N. The trace of A is defined as the sum of the elements on its leading diagonal. Denoting half-trace of A by α1 , we have 2α1 = a11 + a22 + · · · + aNN

(8.156)

The minor Mij of matrix A is the determinant of the N − 1 by N − 1 matrix obtained by deleting row i and column j from the original matrix. A minor is called a principal minor when i = j.

Analysis of single-degree-of-freedom systems 383

The eigenvalues λ of a square matrix A are obtained by solving the characteristic polynomial equation det [A − λI] = 0

(8.157)

In particular, when A is a 3 × 3 matrix, expansion of Equation 8.157 gives λ3 − 2α1 λ2 + a2 λ − a3 = 0

(8.158)

where α2 is the sum of principal minors of A and α3 is the determinant of A. Solution of Equation 8.158 will give three different values of λ. The three eigenvalues are denoted by λ1 , λ2 , and λ3 . The spectral radius, ρ(A), of matrix A is defined by ρ(A) = max{|λ1 |, |λ2 |, |λ3 |} We examine the stability of numerical integration technique with reference to the solution of the equation of undamped free vibration (Eq. 8.153). The condition of stability will be affected by the presence of damping, but for small amounts of damping this effect can be ignored. In any case, the presence of damping will make the stability conditions less restrictive. If a solution method is unstable under conditions of free vibration, it is also likely to be unstable for the solution of the problems of forced vibration, because instability in the complementary solution will soon make the total solution meaningless. In a majority of numerical integration techniques presented in this chapter, the three response parameters, displacement, velocity, and acceleration, at time point n + 1 can be expressed in terms of their values at point n as follows: rn+1 = Arn

(8.159)

where rnT = [un hu˙ n h2 u¨ n ] and A is called the amplification matrix. Equation 8.159 assumes that the exciting force is zero. We write Equation 8.159 at two previous time points, rn = Arn−1 rn−1 = Arn−2

(8.160)

It can be shown that the use of Equations 8.159 and 8.160 to eliminate the velocities and accelerations gives an equation of the form un+1 − 2α1 un + α2 un−1 − α3 uu−2 = 0

(8.161)

where, as defined earlier, α1 is 12 trace of A, α2 is the sum of principal minors of A, and α3 is the determinant of A. Equation 8.161 is a difference equation which has a solution of the form un = cρ n

(8.162)

384

Dynamics of structures

in which c is a constant. Substitution of Equation 8.162 in Equation 8.161 gives ρ 3 − 2α1 ρ 2 + α2 ρ − α3 = 0

(8.163)

Equation 8.163 has three roots ρ1 , ρ2 , and ρ3 . By comparing Equations 8.163 and 8.158, we can conclude that these roots are, in fact, equal to the three eigenvalues of matrix A. A general solution of Equation 8.161 is given by un =

3

ci ρin

(8.164)

i=1

where ci ’s are arbitrary constants to be determined from the initial conditions. If the absolute value of any of the three roots of Equation 8.163 is greater than 1, the solution given by Equation 8.164 will become unbounded as time progresses, that is, as n increases. In such a situation, the solution procedure is said to be unstable. Stability thus requires that the absolute value of each of the three roots of Equation 8.163 be less than or equal to 1, or equivalently that the spectral radius of matrix A be less than or equal to 1. Comparison between Equations 8.164 and 8.154 shows that the difference equation has three roots, while the differential equation has only two. The extra root, denoted by ρ3 , is called a spurious root, while the other two roots, ρ1 and ρ2 , are the principal roots. If the contribution from the spurious root has to be small, it must be real and less than 1. Also, the differential equation has an oscillating solution; therefore, if the solution of the difference equation has to correspond to that of the differential equation, the two principal roots ρ1 and ρ2 must be complex conjugates of each other. In the following we examine the stability of each of the integration methods described in this chapter.

8.16.1 Newmark’s β method The amplification matrix for Newmark’s β method is obtained by simultaneous solution of Equations 8.108 and 8.114 and the equation of motion at time point n + 1 for the three parameters un+1 , u˙ n+1 , and u¨ n+1 . After some algebraic manipulation, we obtain  A=

1 + 2ξ γ ωh

1  −γ ω2 h2 D −ω2 h2

1 + 2ξ ωh(γ − β) 1 + ω2 h2 (β − γ ) −(2ξ ωh + ω2 h2 )

where D = 1 + 2ξ γ ωh + βω2 h2

1

    − β + 2ξ ωh 12 γ − β    (8.165) (1 − γ ) + ω2 h2 β − γ2    2 2 1 −2ξ ωh(1 − γ ) − ω h 2 − β 2

Analysis of single-degree-of-freedom systems 385

With ξ = 0 the invariants of matrix A, α1 , α2 , and α3 become   1 + ω2 h2 /2 − 12 + 2β − γ α1 = 1 + βω2 h2   1 + ω2 h2 12 + β − γ α2 = 1 + βω2 h2 α3 = 0

(8.166)

Equation 8.163 now reduces to ρ 2 − 2α1 ρ + α2 = 0

(8.167)

where α1 and α2 are given by Equation 8.166. In fact, Equation 8.167 is, in this case, obtained more easily directly from Equations 8.108 and 8.114. We begin by rewriting the velocity equation, Equation 8.108, at time tn = nh: u˙ − u˙ n−1 = h(1 − γ )u¨ n−1 + γ hu¨ n

(8.168)

Next we write the displacement equation (Eq. 8.114) at tn :
un = un−1 + hu˙ n−1 + h

2

 1 − β u¨ n−1 + h2 β u¨ n 2

(8.169)

Subtraction of Equation 8.169 from Equation 8.114 without the remainder term gives un+1 −2un +un−1


 1 = h(u˙ n − u˙ n−1 )+h − β (u¨ n − u¨ n−1 )+h2 β(u¨ n+1 − u¨ n ) (8.170) 2 2

On substituting for u˙ − u˙ n−1 from Equation 8.168 and using Equation 8.153 to express the accelerations in terms of displacements, we get     2 + ω2 h2 − 12 + 2β − γ 1 + ω2 h2 12 + β − γ un+1 − un + un−1 = 0 1 + βω2 h2 1 + βω2 h2

(8.171)

which is the same as Equation 8.167. Equation 8.167 indicates that Newmark’s β method provides only two roots. They are given by ρ1,2 = α1 ±

*

α12 − α2

(8.172)

For the difference equation to provide an oscillatory solution, α12 must be less than α2 . Assuming that this condition is satisfied, Equation 8.172 can be expressed as 1/2

ρ1,2 = α2 (cos φ ± i sin φ)

(8.173)

386

Dynamics of structures

where % tan φ = ±

α2 − α12 α1

(8.174)

The general solution of the difference equation thus becomes n/2

un = α2 (c1 cos nφ + c2 sin nφ)

(8.175)

where c1 and c2 are arbitrary constants to be obtained from initial conditions. The solution given by Equation 8.175 will remain bounded provided that α2 is less than or equal to 1. In summary, then, the condition for stable oscillatory response is α12 < α2 ≤ 1

(8.176)

To obtain the period of oscillation, we rewrite Equation 8.175 in the form n/2

¯ n + c2 sin ωt ¯ n) un = α2 (c1 cos ωt

(8.177)

¯ This will, where tn = nh and ω¯ = φ/h. The period of oscillation is given by T¯ = 2π/ω. in general, be slightly different from the true period 2π/ω. Constant-acceleration method In this case we have γ = 0 and β = 0, so that α1 = 1 −

ω 2 h2 4

ω 2 h2 α2 = 1 + 2

(8.178)

Parameter α2 is always greater than 1 and hence the solution will diverge with time. The response will be oscillatory provided that α12 < α2

(8.179a)

ωh < 4

(8.179b)

or

Average acceleration method In this case, γ = 12 and β = 14 . Hence α1 =

1 − ω2 h2 /4 1 + ω2 h2 /4

α2 = 1

(8.180)

Analysis of single-degree-of-freedom systems 387

Since α2 is equal to 1, the amplitude of the numerical solution is exactly equal to that of the exact solution and does not decay or diverge with time. Also α12 is less than α2 for all values of ωh. The method is therefore unconditionally stable. Linear acceleration method For this method, γ = 12 and β = 16 , so that α1 =

1 − ω2 h2 /3 1 + ω2 h2 /6

(8.181)

α2 = 1

Again, since α2 = 1, the amplitude of the solution is exact. The condition for oscillatory motion gives ωh
1.37, ρ(A) is always less than 1. Wilson-θ method can therefore be made unconditionally stable by selecting θ > 1.37. As we shell see later, the accuracy of the method is better for smaller θ and a value θ = 1.4 is usually selected. We now examine the three roots of the difference equation (Eq. 8.163) for two different values of θ : 1.4 and 2.0. The roots can be obtained by solving the cubic equation, or equivalently, by calculating the eigenvalues of amplification matrix A (Eq. 8.183) with ξ = 0. The absolute values of the three roots are plotted in Figure 8.14 as functions of h/T = ωh/2π . It is found that one of the three roots is real. This root represents a spurious root in the solution. Its value is always less than 1, and hence its contribution dies out fairly rapidly. The other two roots, called the principal roots, are complex conjugates of each other, so that their absolute values are the same. This absolute value is found to be less than 1, which implies that the amplitude of calculated response decays with time and the numerical process has introduced an artificial damping.

Analysis of single-degree-of-freedom systems 389

Figure 8.13 Wilson-θ method; spectral radius versus θ for h/T = ∞.

Figure 8.14 Roots of the difference equation.

For θ = 1.4, the curve representing the principal roots dips to a minimum at h/T ≈ 2.29. When h/T > 2.29, the principal roots take two distinct, real values and the solution of the difference equation is no longer oscillatory, although it is still bounded. For h/T > 2.29, the curve in Figure 8.14 shows the principal root having the higher absolute value.

390

Dynamics of structures

8.16.3 Central difference method In this case the difference equation is obtained directly from Equation 8.144. With c = 0 and p = 0, we have un+1 − (2 − ω2 h2 )un + un−1 = 0 α1 = 1 −

(8.186)

ω 2 h2 2

(8.187)

α2 = 1

Since α2 is equal to 1, the amplitude of the numerical solution is equal to that of the true solution. The condition for oscillatory motion gives ωh < 2

(8.188a)

h 1 < ≡ 0.318 π T

(8.188b)

or

The spectral radius of the amplification matrix is plotted in Figure 8.14 as a function of h/T. The spectral radius is greater than 1 when h/T > 1/π , confirming that Equation 8.188b defines the condition of stability.

8.16.4 Houbolt’s method The difference equation for free undamped vibration is obtained from Equation 8.152 with pn+1 and c equal to zero: (2 + ω2 h2 )un+1 − 5un + 4un−1 − un−2 = 0

(8.189)

The characteristic equation is (2 + ω2 h2 )ρ 3 − 5ρ 2 + 4ρ − 1 = 0

(8.190)

From Equation 8.190 it is evident that, as in the case of the Wilson-θ method, Houbolt’s method leads to three roots for the characteristic equation. Of these, the spurious root is found to be real, while the other two principal roots are complex conjugates of each other. The absolute values of the spurious root and the principal roots are plotted in Figure 8.14 as functions of h/T. These values are less than 1 for all h/T and the method is therefore unconditionally stable.

8.17

SELECTION OF A NUMERICAL INTEGRATION METHOD

A number of alternative methods of numerical integration have been presented in the preceding sections. It is observed that with the exception of constant-acceleration

Analysis of single-degree-of-freedom systems 391

method, all of these methods are either unconditionally stable or stable when ωh or h/T is below a certain value. In all cases where there is a condition, the limiting value of ωh is seen to be larger than 1. On the other hand, as pointed out earlier, to ensure adequate accuracy in the numerical results, ωh must be selected to be considerably smaller than 1. It can therefore be concluded that for the response analysis of a singledegree-of-freedom system, the stability criteria are not restrictive and the selection of the method is not therefore governed by considerations of stability. As we shall see later, the situation is different for multi-degree-of-freedom systems where stability may be an important consideration. Stability not being a criterion, the selection of numerical integration must be based on the relative accuracy of the results obtained. We measure the relative accuracy in reference to the solution of the undamped free-vibration equation (Eq. 8.153), which has a solution given by Equation 8.154. The numerical integration procedures, on the other hand, give a solution of the form un = c3 ρ3n + c1 ρ1n + c2 ρ2n

(8.191)

where ρ3 is a spurious root, and ρ1 , ρ2 are the principal roots which are of the form a + bi and a − bi. Equation 8.191 can be expressed in the alternative form un = c3 ρ3n + ρpn (ˆc1 cos ωt ¯ n + cˆ 2 sin ωt ¯ n)

(8.192)

√ where ρp = a2 + b2 , tan φ = b/a, tn = nh, ω¯ = φ/h, and cˆ 1 and cˆ 2 are new arbitrary constants. In some of the integration methods presented here, the spurious solution does not exist. When it does exist, it rapidly dies out because ρ3 is less than 1. We therefore restrict our comparison to the principal part of the solution. If the true solution (Eq. 8.154) and the numerical solution have to correspond to each other, ρp must be equal to 1 and ω¯ must be equal to ω. In general, these conditions are not satisfied. As measures of relative accuracy, we therefore define a period elongation (PE) and an amplitude decay (AD) as follows: PE =

T¯ − T ωh − φ = T φ

(8.193a)

¯ n+T/h

AD =

ρpn − ρp ρpn

2π/φ

= 1 − ρp

(8.193b)

where T¯ = 2π/ω. ¯ As an alternative, amplitude decay can also be represented by defining an equivalent viscous damping coefficient ξ¯ such that ¯

¯ n ρpn = e−ξ ωt ¯

= e−ξ φn

(8.194)

392

Dynamics of structures

Figure 8.15 Period elongation versus h/T.

Equation 8.194 gives ξ¯ = −

ln ρp φ

(8.195)

The period elongation for several different methods is plotted in Figure 8.15 as a function of h/T. In Figure 8.16, we show amplitude decay as well as the equivalent viscous damping as a function of h/T for both the Wilson-θ method and Houbolt’s method. It should be noted that the average acceleration method and the linear acceleration method show no amplitude decay. From the data presented in Figures 8.15 and 8.16, it is evident that for the numerical integration of single-degree-of-freedom systems, the linear acceleration method,

Analysis of single-degree-of-freedom systems 393

Figure 8.16 Amplitude decay versus h/T.

which gives no amplitude decay and the lowest period elongation, is the most suitable of the methods presented.

8.18

SELECTION OF TIME STEP

The selection of an appropriate time step for the numerical integration of the equation of motion is important in the success of the procedure. Too large a time step will give results that are unsatisfactory from the point of view accuracy. A very small time step, on the other hand, will mean increased cost of computation. The natural period of the system as well as the characteristics of the exciting force are factors that govern the choice of a time step. An examination of the truncation error terms for the methods presented in the preceding sections shows that to keep the errors within limits, ωh must be less than 1. This is equivalent to h/T being less than 0.16. In general, acceptable accuracy can be achieved with a time step that is about one-tenth of the natural period of the system.

394

Dynamics of structures

Numerical integration involves sampling of the exciting function at intervals equal to the selected time step. Such sampling should not lead to a distortion of the exciting function. If the function is viewed as a superposition of harmonic components, a time step that is about one-tenth the component with the smallest period should be quite satisfactory. It is not, however, necessary to decompose the forcing function into its harmonic components to determine a suitable value of h. In general, the time step size can be determined by an inspection of the exciting function. Further, the step size need not be much smaller than that governed by the natural period of the system, because high-frequency components of the exciting force which are not adequately represented by such a selection will not, in any case, excite a significant dynamic response of the system. The accuracy of the numerical integration process may be verified by performing the response calculations with two different time steps that are close to each other. If the response obtained with the shorter time step is not too different from that obtained with the larger time step, the process may be taken to have converged to the true solution. SELECTED READINGS Argyris, J.H., Dunne, P.C., and Angelopoulos, T., “Dynamic Response by Large Step Integration’’, Earthquake Engineering and Structural Dynamics, Vol. 2, 1973, pp. 185–203. Bathe, K.J., Finite Element Procedures, Prentice-Hall, Englewood Cliffs, 1996. Bathe, K. J., and Wilson, E. L., “Stability and Accuracy Analysis of Direct Integration Methods,’’ Earthquake Engineering and Structural Dynamics, Vol. 1, 1973, pp. 283–291. Belytschko, T. and Scoeberle, D.F., “On the Conditional Stability of an Implicit Algorithm for Nonlinear Structural Dynamics’’, Journal of Applied Mechanics, Vol. 42, 1975, pp. 865–869. Clough, R.W. and Penzien, J., Dynamics of Structures, 2nd Edition, McGraw-Hill, New York, 1993. Goudreau, G.L. and Taylor, R.L., “Evaluation of Numerical Integration Methods in Elastodynamics’’, Computer Methods in Applied Mechanics and Engineering, Vol. 2, 1972, 69–97. Hamming, R.W., Numerical Methods for Scientists and Engineers, McGraw-Hill, New York, 1962. Hilber, H.M., and Hughes, T.J.R., “Collocation, Dissipation, and Overshoot for Time Integration Schemes in Structural Dynamics,’’ Earthquake Engineering and Structural Dynamics, Vol. 6, 1978, pp. 99–117. Hilber, H.M., Hughes, T.J.R., and Taylor, R.L., “Improved Numerical Dissipation for Time Integration Algorithms in Structural Dynamics,’’ Earthquake Engineering and Structural Dynamics, Vol. 5, 1977, pp. 283–292. Houbolt, J.C., “A Recurrence Matrix Solution for the Dynamic Response of Elastic Aircraft,’’ Journal of the Aeronautical Sciences, Vol. 17, 1950, pp. 540–550. Hughes, T.J.R., “Stability, Convergence and Growth and Decay of Energy of the Average Acceleration Method in Nonlinear Structural Dynamics’’, Computers and Structures, Vol. 6, 1976, pp. 313–324. Humar, J. L., and Wright, E.W., “Numerical Methods in Structural Dynamics,’’ Canadian Journal of Civil Engineering, Vol. 1, 1974, pp. 179–193. Hurty, W.C., and Rubinstein, M.F., Dynamics of Structures, Prentice Hall, Englewood Cliffs, 1964. Krieg, R.D., “Unconditional Stability in Numerical Time Integration Methods’’, Journal of Applied Mechanics, Vol. 40, 1973, pp. 417–421.

Analysis of single-degree-of-freedom systems 395 Mewmark, N.M., “A Method of Computation for Structural Dynamics,’’ Journal of Engineering Mechanics Division, ASCE, Vol. 85, 1959, pp. 67–94. Rayleigh, Lord, The Theory of Sound, Dover, New York, 1945. Temple, G., and Bickley, W.G., Rayleigh Principle and its Application to Engineering, New York, 1956. Thomson, W.T., Theory of Vibration with Applications, 2nd Edition, Prentice Hall, Englewood Cliffs, 1964.

PROBLEMS 8.1

Obtain the frequency of vibration of the system shown in Problem 2.1 by using Rayleigh’s method.

8.2

Obtain the frequency of vibration for small oscillations of the system shown in Problem 2.3. Neglect damping and use Rayleigh’s method.

8.3

Using Rayleigh’s method, obtain the frequency of vibration for small oscillations of the inverted pendulum shown in Figure P8.3.

Figure P8.3 8.4

A steel beam is simply supported over a span of 4.0 m carries weights of 20 kN at the center and 12 kN at 1.0 m from each end (Fig. P8.4). Calculate the frequency of transverse vibration assuming that the deflection shape is similar to the static deflection shape and that the mass of the beam can be neglected in comparison to the mass of the concentrated weights. The static deflection is 1.2 mm at the 20-kN load and 0.85 mm at each of the 12-kN loads.

Figure P8.4

396 8.5

Dynamics of structures The aluminum beam shown in Figure P8.5 carries a small hoist weighing 450 N. Calculate the frequency of vibration when the hoist is at a distance of 1.2 m. from one support. What are the lowest and highest frequencies and the corresponding hoist locations? E = 69,000 MPa and ρ = 2770 kg/m3 for aluminum.

Figure P8.5 8.6

Obtain the frequency of vibration of the structure shown in Figure P8.6 by using Rayleigh’s method. Neglect deformations due to shear and axial forces and assume that the vibration shape is that produced by a vertical downward load acting at point C.

Figure P8.6 8.7

Determine the frequency of vibration of the beam shown in Figure P8.7 by using Rayleigh’s method. Assume that the vibration shape is similar to that produced by the deflection of the beam under self weight gravity load applied in the appropriate direction.

Figure P8.7

Analysis of single-degree-of-freedom systems 397 8.8

Obtain an estimate of the fundamental frequency of the two degree-of-freedom system shown in Figure P8.8 using Rayleigh’s method and the following deflection shape. 

u10 ψ0 = u20



  1 = 1

Refine your frequency estimate by using the improved Rayleigh’s method.

Figure P8.8 8.9

The single story building shown in Figure P8.9 is subjected to a blast load whose effect can be represented by a story level lateral force as indicated. The total lateral stiffness of the building is 14 kips/in and the floor deck mass is 0.1 kip · s2 /in · Obtain the displacement

Figure P8.9

398

Dynamics of structures response history for the first 0.5 s by a numerical integration of the Duhamel’s integral. Use a time step of 0.05 s and the trapezoidal method of integration.

8.10

Obtain the response of the structure in Problem 8.9 if damping is 10% of critical. As in Problem 8.9, evaluate the Duhamel’s integral by the trapezoidal method of integration and use a time step of 0.05 s.

8.11

Obtain the response of the structure in Problem 8.9 by the average acceleration method. Assume that damping is 10% of critical and use a time step of 0.05 s.

8.12

Repeat Problem 8.11 using the linear acceleration method.

8.13

Repeat Problem 8.11 using the central difference method.

8.14

Solve Problem 8.11 by Houbolt’s method. To start the numerical integration use the displacement values at 0.05 s and 0.10 s obtained in Problem 8.11.

8.15

An eccentric mass shaker installed on the top floor of the three-story shear frame building shown in Figure P8.15 produces a harmonic force p0 sin t, where p0 = 45 kN and  = 5π/3 rad/s. The frame is modeled as a single-degree-of-freedom system by assuming that the vibration shape is ψ T = [1 1.75 2.5]. Find the history of the top-floor displacement for the first 0.6 s using a piece-wise linear representation of the forcing function and a time step of 0.1 s. Find the maximum base shear produced during this time. Neglect damping.

Figure P8.15 8.16

Repeat Problem 8.15 assuming that damping is 10% of critical.

Chapter 9

Analysis of response in the frequency domain

9.1 TRANSFORM METHODS OF ANALYSIS The methods of time-domain analysis outlined in Chapters 5 through 8 can be effectively used to obtain the response of any single-degree-of-freedom system. It is, however, possible in certain situations to greatly improve the efficiency of the analytical procedure by employing the transform method of mathematics. Transform methods have proved to be very useful in many engineering applications. The simplest of these methods is the logarithm. Logarithms allow the transformation of a problem of algebraic multiplication to a simpler problem of addition. The success of the method, however, depends on the availability of the tables of logarithms and antilogarithms of numbers. Logarithm is a transform that applies to discrete value of a variable. There are other transforms that apply to continuous functions; Laplace transform is one of them. It allows the conversion of a differential equation to an algebraic equation which is much simpler to solve. Again, the success of the method depends on the availability of the tables of Laplace transforms and inverse Laplace transforms. In the analysis of dynamic response of linear structures, Fourier transform can play an important role. Briefly, this transform converts the task of evaluating the convolution integral of Equation 7.9 into that of finding the product of the Fourier transforms of the two functions involved in the convolution. The inverse Fourier transform of the product then gives the desired response function of time. The method of analysis is commonly referred to as dynamic analysis in the frequency domain. It would appear that like other transforms, the success of the method should depend on the ease with which the Fourier transform and the inverse Fourier transform of the functions involved in the analysis can be evaluated. Unfortunately, except for simple cases, where the direct method of solving the convolution integral is as effective, the application of Fourier transform involves evaluation of integrals that are not always easy to solve. In most practical cases, therefore, Fourier transform is used in its discrete form, which permits the replacement of the integrals by equivalent numerical computations of areas. In fact, the efficiency of Fourier methods is most apparent in cases where the excitation is specified in terms of numerical values at regular intervals of time rather than as a mathematical function. In such a situation, even the time-domain analysis, which eventually depends on the evaluation of a convolution integral, must rely on

400

Dynamics of structures

numerical methods, and the alternative method of analysis through discrete Fourier transform proves to be most effective. Ordinarily, the computations involved in obtaining the discrete Fourier transforms of the functions being convolved, taking the product of these transforms and then evaluating the discrete inverse Fourier transform are no less than those in a direct evaluation of the discrete convolution. However, the development of a special algorithm called fast Fourier transform (FFT) has completely altered this position. The FFT algorithm, which derives its efficiency from exploiting the harmonic property of a discrete transform, cuts down the computations by several orders of magnitude and makes frequency-domain analysis highly efficient. The FFT has found widespread application in several areas of engineering analysis. Its use in dynamic analysis of structures is, however, more recent. Textbooks on structural dynamics therefore contain at best a very cursory treatment of the topic. The subject is, however, important enough to justify a more in-depth coverage. Besides computational efficiency, analysis in the frequency domain possesses several other advantages. By providing a clearer representation of the frequency content of the forcing function, it enables one to evaluate the potential for it to excite a given structure. In addition, for problems involving infinite domains such as those of wave motion in reservoirs of large extent or in unbounded soil media, and in other situations where the physical characteristics of the system are dependent on the vibration frequency, frequency-domain analysis may prove to be more effective than a timedomain analysis. It should, however, be noted that frequency-domain analysis applies only to time-invariant, linear systems for which the principle of superposition holds. This chapter begins by presenting the response analysis for periodic excitation, which, in fact, is an example of the simplest type of frequency-domain analysis. Analytical methods for evaluating the response of nonperiodic excitation are dealt with next. The concept of Fourier transforms is developed at this stage. This is followed by the introduction of discrete Fourier transforms and their application in the evaluation of dynamic response.

9.2

FOURIER SERIES REPRESE NTATION OF A PERIODIC FUNCTION

A function g(t) of time t is said to be a periodic function of t with a period equal to T0 if it satisfies the following relationship g(t + nT0 ) = g(t)

(9.1)

where n is an integer with negative or positive values. Figure 9.1 presents several examples of periodic functions. A periodic function with a finite number of discontinuities and a finite number of maximas or minimas within a range of time equal to its period T0 can be represented by an infinite trigonometric series as follows g(t) = a0 +

∞

n=1

an cos 2πnf0 t +

∞

n=1

bn sin 2π nf0 t

(9.2)

Analysis of response in the frequency domain 401

Figure 9.1 Examples of periodic functions.

where an and bn are constants to be determined and f0 = 1/T0 is the frequency in cycles per second. The trigonometric series of Equation 9.2 is known as the Fourier series. Setting ω0 = 2π f0 , where ω0 is the frequency in radians per second, the series can be expressed in the alternative form g(t) = a0 +

∞

an cos nω0 t +

n=1

∞

bn sin nω0 t

(9.3)

n=1

To obtain the coefficients an , we multiply the left- and the right-hand sides of Equation 9.3 by cos mω0 t and integrate over a period. Then noting that   T0 /2 0 for n = m (9.4a) cos nω0 t cos mω0 t dt = T0 /2 for n = m −T0 /2 

T0 /2

−T0 /2

 cos mω0 t dt =

0 T0

for m = 0 for m = 0

(9.4b)

402

Dynamics of structures

and 

T0 /2

−T0 /2

sin nω0 t cos mω0 t dt = 0

for all m and n

(9.4c)

we obtain the following expression for coefficients a: a0 =

1 T0

an =

2 T0



T0 /2

−T0 /2



g(t) dt

(9.5a)

g(t) cos nω0 t dt

(9.5b)

T0 /2

−T0 /2

Coefficients bn are obtained in a similar manner by multiplying the left- and the righthand sides of Equation 9.3 by sin mω0 t and integrating over a period. Then, because of the relationships 

T0 /2

−T0 /2

 sin nω0 t sin mω0 t dt =

0 for n = m T0 /2 for n = m

(9.6)

and that given by Equation 9.4c, the coefficient bn is obtained as bn =

2 T0



T0 /2

−T0 /2

g(t) sin nω0 t dt

(9.7)

It can be shown that at a discontinuity such as the one shown in Figure 9.1b, the Fourier series of Equation 9.2 or 9.3 converges to a value that is the average of the values of the function g(t) immediately to the left and the right of the discontinuity.

9.3

RESPONSE TO A PERIODICALLY APPLIED LOAD

From the foregoing discussion we note that any periodic load can be expressed in terms of the sum of a series of harmonic components. Theoretically speaking, an infinite number of such components must be included to get a precise representation of the load. In practice, the contribution of the successive terms becomes smaller and smaller, and only the first few terms need to be included in the series to obtain a reasonable level of accuracy. Because each component in the series represents a harmonic load, the response to which we already know how to evaluate, the total response of a singledegree-of-freedom system to the periodic load can be obtained simply by summing the responses to the individual components provided that the system is linear and the principle of superposition holds. It should be noted, however, that our periodic representation of the load necessarily implies that the load has been in existence for an indefinite period of time even before time zero. The response that we can hope to evaluate from such a periodic representation will therefore be the steady-state response to the applied load and will not include the transient term. In Chapter 6 we developed expressions for the steady-state response of a system to several different types of applied loads. When the applied load is constant and has

Analysis of response in the frequency domain 403

a magnitude p0 , the steady-state response is given by u(t) =

p0 k

(9.8)

For a sine function load, p0 sin t, the steady-state response of an undamped system is u(t) =

p0 1 sin t k 1 − β2

(9.9)

where β = /ω and ω is the natural frequency of the system. The response of a damped system is given by u(t) =

p0 1 (1 − β 2 ) sin t − 2ξβ cos t 2 2 2 k (1 − β ) + (2ξβ)

(9.10)

For a cosine function load, p0 cos t, the undamped response is u(t) =

p0 1 cos t k 1 − β2

(9.11)

while the damped response is u(t) =

p0 1 2ξβ sin t + (1 − β 2 ) cos t 2 2 2 k (1 − β ) + (2ξβ)

(9.12)

Using expressions 9.8, 9.9, and 9.11, the undamped steady-state response of a system to a periodic force represented by the Fourier series of Equation 9.3 is obtained as u(t) =

∞

∞

n=1

n=1

bn 1 a0 an 1 + cos nω0 t + sin nω0 t 2 k k 1 − βn k 1 − βn2

(9.13)

where βn = nω0 /ω. In a similar manner, the damped response of the system can be shown to be ∞

u(t) =

a 0 an 1 + {2ξβn sin nω0 t + (1 − βn2 ) cos nω0 t} 2 2 k k (1 − βn ) + (2ξβn )2 n=1

+

∞

n=1

bn 1 {(1 − βn2 ) sin nω0 t − 2ξβn cos nω0 t} 2 2 k (1 − βn ) + (2ξβn )2 ∞

=

1 a0 1 [{an 2ξβn + bn (1 − βn2 )} sin nω0 t + 2 2 k (1 − βn ) + (2ξβn )2 k n=1

+ {an (1 − βn )2 − bn 2ξβn } cos nω0 t] (9.14) Equations 9.13 and 9.14 can be recognized as Fourier series representations of the response, which must therefore be periodic with a period equal to T0 .

404

Dynamics of structures

Example 9.1 Obtain the steady-state response of the system shown in Figure E9.1a to the loading shown in Figure E9.1b.

Solution To obtain a periodic representation of the excitation force, we extend the function on the negative direction of the time axis as shown in Figure E9.1b. The period T0 is seen to be 4 s and the coefficients of the Fourier series are therefore given by a0 =

1 4

an =

2 4

bn =

2 4



2

g(t) dt

(a)

−2



2

g(t) cos

2nπ t dt 4

(b)

g(t) sin

2nπt dt 4

(c)

−2



2

−2

Figure E9.1 Response of a single-degree-of-freedom system to a periodic load: (a) system model; (b) periodic load; (c) first three harmonics of the loading function; (d) Fourier series representation of the loading function; (e) response to periodic load.

Analysis of response in the frequency domain 405 Because g(t) is even, Equations a, b, and c give a0 =

1 2



2

g(t) dt 0

 1  2 1 1 p0 dt − p0 dt = 0 2 2 0 1  2 2nπt dt g(t) cos an = 4 0  1  2 2nπt 2nπt dt − p0 dt = p0 cos cos 4 4 0 1 =

(d)

nπ 4p0 sin 2 nπ bn = 0 =

(e) (f)

The Fourier series representation of the periodic forcing function is, therefore, 4p0 g(t) = π




πt 1 3πt 1 5πt 1 7πt cos − cos + cos − cos + ··· 2 3 2 5 2 7 2

 (g)

The first three harmonics and their sum are shown in Figure E9.1c and d. It is seen that the first harmonic term, which has a period of 4 s or a frequency of 14 Hz, has an amplitude of 4p0 /π. The amplitude decreases rapidly as the frequency increases, so that the second harmonic with a frequency of 34 Hz has an amplitude of only 4p0 /3π, while the third harmonic has a frequency of 54 Hz and an amplitude of 4p0 /5π . It is also seen that even when just three terms are included, the series provides a reasonable representation of the rectangular forcing function. At t = 1 s where the forcing function is discontinuous, the Fourier series converges to the average value of the function, which in this case is zero. The natural frequency of the system is ω = k/m = π rad/s. The response of the system, which is undamped, is now obtained by substituting the value of an in Equation 9.13 u(t) =

4p0 πk




1 1 1 1 1 πt 3π t 5πt − + + ··· cos cos cos 2 3 1 − β32 2 5 1 − β52 2 1 − β12

 (h)

where βn = nω0 /ω = n/2. Figure E9.1e shows the response over half a period, obtained by summing the first three terms of the series in Equation h.

9.4

EXPONENTIAL FORM OF FOURIER SERIE S

In developing the frequency-domain analysis procedure, it is convenient first to obtain an exponential form for the Fourier series of Equation 9.3. Using de Moivre’s theorem, the sine and cosine functions in Equation 9.3 can be expressed in terms of complex exponentials as follows sin nω0 t =

einω0 t − e−inω0 t 2i

(9.15)

406

Dynamics of structures

einω0 t + e−inω0 t 2

cos nω0 t =

(9.16)

Substitution of Equation 9.16 in Equation 9.5b yields 1 an = T0



T0 /2

g(t)(einω0 t + e−inω0 t ) dt

−T0 /2

(9.17a)

If we replace n by −n in Equation 9.17a, we get a−n

1 = T0



T0 /2

−T0 /2

g(t)(e−inω0 t + einω0 t ) dt = an

(9.17b)

In a similar manner, bn is obtained by substituting Equation 9.15 in Equation 9.7: bn = b−n

1 T0

1 = T0



g(t)

einω0 t − e−inω0 t dt i

(9.18a)

g(t)

e−inω0 t − einω0 t dt = −bn i

(9.18b)

T0 /2

−T0 /2



T0 /2

−T0 /2

The Fourier series of Equation 9.3 can be expressed as g(t) = a0 +

∞

n=1

= a0 +

∞

∞

an

einω0 t + e−inω0 t eiω0 nt − e−inω0 t bn + 2 2i


einω0 t

n=1

an bn + 2 2i

n=1

 +

∞

e−inω0 t




n=1

an bn − 2 2i

 (9.19)

Using relationships 9.17b and 9.18b, Equation 9.19 becomes g(t) = a0 +

∞


einω0 t

n=1

=

∞

an bn + 2 2i

 +

−1

n=−∞


einω0 t

an bn + 2 2i



cn einω0 t

(9.20)

n=−∞

in which coefficient cn is given by 1 (an − ibn ) 2  T0 /2

1 = g(t) (einω0 t + e−inω0 t ) − (einω0 t − e−inω0 t ) dt 2T0 −T0 /2  T0 /2 1 g(t)e−inω0 t dt = T0 −T0 /2

cn =

(9.21a)

Analysis of response in the frequency domain 407

and 1 c0 = T0



T0 /2

−T0 /2

g(t) dt = a0

(9.21b)

Equations 9.20 and 9.21 together represent the exponential form of the Fourier series, which is seen to be much more compact than the alternative form given by Equations 9.3, 9.5, and 9.7.

9.5

COMPLEX FREQUENCY RESPONSE FUNCTION

In Section 9.3 we developed a method for obtaining the response of a system to a periodically applied load. The procedure we used was first to express the applied load as the sum of a series of harmonic components. Knowing the steady-state response to a harmonic load, we could obtain the response to each harmonic component of the periodic load. The sum of these component responses gave us the total response. We can use a similar procedure when the Fourier series representing the applied load is given in the exponential form, but we must first obtain the steady-state response of the system to an excitation given by the complex function p = p0 eiω0 t . This can be achieved by solving the following differential equation for u. mu¨ + cu˙ + ku = p0 eiω0 t

(9.22)

The complementary solution to Equation 9.22 is given by Equation 5.36. u = e−ξ ωt ( A cos ωd t + B sin ωd t )

(5.36)

The particular integral solution of Equation 9.22 is of the form u = Geiω0 t

(9.23)

Substituting u and its derivatives from Equation 9.23 into Equation 9.22, we get (−mω02 + icω0 + k)Geiω0 t = p0 eiω0 t

(9.24)

The complete solution of Equation 9.22 is then obtained by summing Equations 5.36 and 9.23 and substituting the value of G from Equation 9.24: u = e−ξ ωt (A cos ωd t + B sin ωd t) +

−mω02

p0 eiω0 t + icω0 + k

(9.25)

The first term on the right-hand side of Equation 9.25 rapidly dies out with time and represents the transient response. The second term in Equation 9.25 is the steadystate response. In summary, the steady-state response of a system to an applied load

408

Dynamics of structures

p0 eiω0 t is given by H(ω0 )p0 eiω0 t , where H(ω0 ) = =

1 −mω02 + icω0 + k k( −

β2

1 + 2iξβ + 1)

(9.26a) (9.26b)

Function H(ω0 ) is called the complex frequency function. When damping is zero, Equation 9.26 reduces to H(ω0 ) =

1 − ω02 )

m(ω2

(9.27)

The response to a periodic load represented by Equation 9.20 is now easily obtained as u(t) =

∞

cn H(nω0 )einω0 t

(9.28a)

n=−∞

where cn =

9.6

1 T0



T0 /2

−T0 /2

g(t)e−inω0 t dt

(9.28b)

FOURIER INTEGRAL REPRESENTATION OF A NONPERIODIC LOAD

To obtain a Fourier representation of a nonperiodic load, we first construct a periodic version of the given load. The first step is to select a value for the period. The selected period should, of course, be larger than the duration of the applied load. Within each period, the periodic version has a magnitude equal to the specified load for the duration of the latter but is zero otherwise. A periodic version constructed as above is shown in Figure 9.2. It is evident that the curves shown by dashed lines represent fictitious loads that, in fact, do not exist. If T0 is now increased to a very large value, the fictitious loads will move to infinity and in the limit we will get a true representation of the applied

Figure 9.2 Periodic version of a loading function.

Analysis of response in the frequency domain 409

load. We apply this reasoning to derive a Fourier representation of the applied load from Equations 9.20 and 9.21. First since T0 is very large, we set ω0 =

2π =  T0

(9.29a)

and nω0 = n

(9.29b)

With this notation, Equation 9.21 gives  cn T0 =

T0 /2

−T0 /2

g(t)e−in t dt

(9.30)

From Equation 9.20, ∞ 1

g(t) = cn T0 ein t T0 −∞ ∞ 1

= cn T0 ein t  2π −∞

(9.31)

In the limit as T0 approaches infinity, n becomes a continuous function and Equation 9.30 becomes  cn T0 = G() =

∞

−∞

g(t)e−it dt

(9.32)

while Equation 9.31 takes the form g(t) =

1 2π



∞

−∞

G()eit d

(9.33)

where we have replaced the discrete coordinate n = n  by its continuous version . Equation 9.32 represents the Fourier transform of the time function g(t), while Equation 9.33 is the inverse Fourier transform. The two together are called Fourier transform pair. They are central to the analysis through frequency domain. In the discussion that follows, we will denote the time function by a lowercase letter, and the Fourier transform of the function by the same uppercase letter. The relationship between a transform pair will be expressed by the following notation g(t) ⇐⇒ G()

(9.34)

410

Dynamics of structures

9.7

RE SPONSE TO A NONPERIODIC LOAD

Having obtained the Fourier representation of a nonperiodic load, it is straightforward to obtain the response of a linear system to an applied load that is nonperiodic. Thus, when T0 is large, Equations 9.28 and 9.29 give u(t) = =

∞ 1

(cn T0 )H(n )ein t T0 −∞ ∞ 1

(cn T0 )H(n )ein t  2π −∞

(9.35)

In the limit as T0 approaches ∞, Equation 9.35 becomes 1 u(t) = 2π



∞

−∞

G()H()eit d

(9.36)

where G() = cn T0 is given by Equation 9.32. A nonperiodic load of special interest is a unit impulse applied at say t = 0. The unit impulse load is most conveniently expressed in terms of the delta function. Thus g(t) = δ(t)

(9.37)

In Chapter 7 we had derived expressions for the response of a system to a unit impulse load. For an undamped and a damped system, respectively, these expressions are given by Equations 7.4 and 7.5. 1 sin ωt mω 1 −ξ ωt h(t) = e sin ωd t mωd

h(t) =

(7.4) (7.5)

The Fourier transform of the unit impulse is  G() =

∞ −∞

δ(t)e−it dt

=1

(9.38)

and the response to a unit impulse is therefore given by h(t) =

1 2π

1 = 2π



∞

−∞ ∞

G()H()eit d



−∞

H()eit d

(9.39)

Analysis of response in the frequency domain 411

Equation 9.39 shows that h(t) is the inverse transform of H() and we have the important relationship h(t) ⇐⇒ H()

(9.40)

The choice of symbol h to represent the unit impulse function and the uppercase version H to represent the complex frequency response function is thus in conformity with our selected notation for a Fourier transform pair.

9.8

CONVOLUTION INTEGRAL AND CONVOLUTION THEOREM

The convolution integral and the convolution theorem which enables us to obtain the Fourier transform of the convolution integral play a central role in the frequencydomain analysis of the dynamic response of structures. In Chapter 7 we indicated the form of convolution integral and proved that it gives the forced response of a linear time invariant system starting from zero initial conditions. In this section we present a formal definition of the convolution integral, demonstrate its physical interpretation, and show how it is possible to obtain its value through Fourier transform analysis. The convolution of two functions g(t) and h(t), denoted as g(t) ∗ h(t), is given by  ∞ g(τ )h(t − τ ) dτ (9.41) u(t) = g(t) ∗ h(t) = −∞

A physical interpretation of the convolution integral can be provided by considering an example. Let it be required to convolve the two functions of time g(τ ) and h(τ ) shown in Figure 9.3a and b, respectively. Function g(τ ) may, for example, represent an exciting force function while h(τ ) may represent a unit impulse response function. For the purposes of demonstration we have kept the two functions simple. Also note that in keeping with our description of g(τ ) and h(τ ) as possible exciting force and unit impulse response, respectively, we have chosen our functions so that they have a null value for τ less than 0. As shown in Figure 9.3c, function h(−τ ) is the mirror image of h(τ ) about the y axis. The process of obtaining h(−τ ) is also referred to as folding of the original function about the y axis. Function h(t−τ ) is simply the function h(−τ ) shifted to the right by time t. As an example, Figure 9.3d shows h(t − τ ) for t = 2. The integrals of the product function g(τ )h(t − τ ) between limits −∞ and ∞ are shown in Figure 9.3g for different values of t. We note that since g(τ ) is 0 for τ < 0, the lower limit of integration is effectively zero. In a similar manner, because h(t − τ ) is 0 for τ > t, the upper limit of integration becomes t. The convolution integral of Equation 9.41 can therefore be expressed as  t u(t) = g(t) ∗ h(t) = g(τ )h(t − τ ) dτ (9.42) 0

Equation 9.42 is precisely the same as the Duhamel’s equation. Therefore, when g(τ ) is the exciting force and h(τ ) the unit impulse response function of a linear time invariant system, the convolution integral g(t) ∗ h(t) gives the response of that system at time t.

412

Dynamics of structures

Figure 9.3 Convolution of functions g(t) and h(t).

The frequency-domain analysis procedure depends on an evaluation of the Fourier transform of the convolution integral (Eq. 9.41 or 9.42). The required transform is given by  ∞ ∞ g(τ )h(t − τ ) dτ e−it dt (9.43) U() = −∞

−∞

Analysis of response in the frequency domain 413

By interchanging the order of integration in Equation 9.43, we obtain  ∞ ∞ h(t − τ )e−it dt g(τ ) dτ U() = −∞

−∞

(9.44)

If we set t − τ = α so that t = α + τ , Equation 9.44 becomes  ∞ ∞ h(α)e−iα dα e−iτ g(τ ) dτ U() =  =

−∞ ∞ −∞

−∞

H()e−iτ g(τ ) dτ

= H() G()

(9.45)

Equation 9.45 proves a theorem of great significance in the frequency-domain analysis. It shows that the Fourier transform of the convolution integral is equal to the product of the Fourier transforms of the functions being convolved. As stated earlier, convolution between the exciting force function and the unit impulse response function gives the resultant response of a linear time invariant system to the exciting force. The Fourier transform of the convolution integral in that case is the product of function G(), the transform of the exciting force, and H(), the transform of the unit impulse function. It was shown that the Fourier transform of the unit impulse function is, in fact, equal to the complex frequency response function given by Equation 9.26. In summary, the following relationships hold g(t) ⇐⇒ G() h(t) ⇐⇒ H()

(9.46)

u(t) = g(t) ∗ h(t) ⇐⇒ G() H() It also follows that u(t) can be obtained by taking the inverse Fourier transform of G()H(). Thus  ∞ 1 G()H()eit d (9.47) u(t) = 2π −∞ This is the same relationship that we obtained directly in Section 9.7.

9.9

DISCRETE FOURIER TRANSFORM

In the foregoing sections we developed the essential steps in the frequency domainanalysis of the dynamic response of structures. These steps can be summarized as follows: 1 2 3

Obtain the Fourier transforms of the excitation function and the unit impulse response function. Take the product of the two transforms obtained in step 1. Take the inverse Fourier transform of the product to obtain the desired response.

414

Dynamics of structures

Figure 9.4 Sampled periodic function.

The success of the procedure therefore depends on being able to obtain the direct and inverse Fourier transforms of the desired functions. However, as noted earlier, the transform expressions involve complex integrals whose evaluation is often quite tedious. In fact, in many problems a numerical evaluation of the transforms may be the only practical method of obtaining a solution. Quite often the forcing function or the unit impulse response function or both are available only in the form of discrete values at specified intervals of time. There are also cases where the Fourier transform of the impulse function is available directly but only in a discrete form. In all such situations, recourse must be had to a numerical evaluation of the transforms involved. In view of these considerations, it is useful to develop the concept of discrete Fourier transform and to establish a relationship between continuous and discrete transforms. Consider the time function shown in Figure 9.4. Let it be sampled at N equal intervals from n = 0 to N − 1 implying that N discrete values of the function are known at time steps 0, t, 2t, . . . , (N − 1)t. It can then be shown that with a suitable choice for the coefficients an , the curve represented by the following series can be made to pass through the given points.

g(kt) =

N−1 1

an e2πikt(n/T0 )  2π n=0

=

1 T0

N−1

an e2πikn/N

(9.48)

n=0

where T0 = Nt and  = 2π/T0 . To obtain the coefficient an , we multiply both sides of Equation 9.48 by e−2π ikm/N and take a sum as k varies from 0 to N − 1. N−1

g(k t)e−2π ikm/N =

N−1 N−1 1

an e2π ikn/N e−2πikm/N T0 n=0 k=0

k=0

=

N−1 N−1 1

an e2π ik(n−m)/N T0 n=0 k=0

(9.49)

Analysis of response in the frequency domain 415

The nth term in the outer sum is given by pn =

N−1 1

an e2πik(n−m)/N T0

(9.50)

k=0

For the case when n = m, Equation 9.50 gives pm =

1 am am N = T0 t

(9.51)

When n = m, Equation 9.50 represents a geometric progression with common ratio r = e2π i(n − m)/N whose sum is given by pn =

1 1 − rN an T0 1−r

1 − e2πi(n−m) 1 an T0 1 − e2πi(n−m)/N =0

=

(9.52)

where the last result is obtained by expanding e2π i(n − m) in the numerator by de Moivre’s theorem and noting that cos 2π(n − m) = 1 and sin 2π (n − m) = 0. Equation 9.49 thus reduces to N−1

g(kt)e−2π ikm/N =

k=0

am t

(9.53a)

or am =

N−1

g(kt)e−2π ikm/N t

(9.53b)

k=0

From Equations 9.48 and 9.53 we obtain the following discrete Fourier transform pair: G(n) =

N−1

g(kt)e−2π ikn/N t

k=0

=

N−1

g(k t)e−iktn t

(9.54a)

N−1 1

G(n)e2πikn/N  2π

(9.54b)

k=0

g(kt) =

n=0

416

Dynamics of structures

where G(n) is equivalent to an . From Equation 9.54b, g{(N + k)t} =

N−1 1

G(n)e2πin(k+N)/N  2π n=0

= =

1 2π 1 2π

N−1

G(n)e2πink/N e2π in 

n=0 N−1

G(n)e2πink/N 

n=0

= g(kt)

(9.55)

The function represented by Equation 9.54b is therefore a periodic extension of the original function we were trying to represent, the period being T0 = Nt. In a similar manner we can show that the discrete Fourier transform given by Equation 9.54a is periodic with a period (N). On comparing Equation 9.54a and its continuous counterpart (Eq. 9.32) we note that the two are equivalent. The integral in the continuous version has been replaced by a rectangular sum over a finite time period Nt in the discrete version. Obviously, if the discrete version has to parallel the continuous version closely, the time step t should be sufficiently small. In a similar manner, a comparison of Equations 9.54b and 9.33 shows that the two are equivalent. Again, integration in the continuous version has been replaced by rectangular sum over a finite frequency band N in the discrete version, and if the two versions have to match each other closely, the frequency step  should be small. An important difference exists between the continuous and discrete Fourier transforms. While the continuous transform provides a true representation of the given function, the discrete transform represents only a periodic version of the function. Within a period T0 = Nt, the periodic version is similar to the continuous version; outside the period the two are quite different, unless the original function also happens to be periodic with a period T0 . 9.10

DISCRETE CONVOLUTION AND DISCRETE CONVOLUTION THEOREM

We define the discrete convolution of two functions as u(kt) =

N−1

g(mt)h{(k − m)t}t

(9.56)

m=0

where both g(t) and h(t) are periodic functions with a period Nt, so that g(m + rN)t = g(mt) h(m + rN)t = h(mt)

r = 0, ±1, ±2, . . .

(9.57)

Analysis of response in the frequency domain 417

On comparing Equations 9.41 and 9.56, we note that the discrete convolution is entirely analogous to continuous convolution. Two important differences, however, exist. First, the discrete convolution is carried out on periodic extensions of the actual functions. This is necessary because the use of discrete Fourier transforms is possible only if the functions being convolved are periodic. Depending on the type of functions being convolved, the convolution of the extended functions may or may not be different from that of the original functions. The second difference is that the integration in a continuous convolution is replaced by rectangular summation in the discrete convolution. If the time interval t is small, the difference between the two is expected to be negligible. Analogous to the case of continuous functions, a relationship exists between the discrete convolution of two periodic functions of time and the product of their discrete Fourier transforms. Mathematically, this relationship can be expressed as g(kt) ∗ h(kt) ⇐⇒ G(n)H(n)

(9.58)

A discrete Fourier transform of the convolution exists only when both g(kt) and h(kt) are periodic. It is then obtained by using Equation 9.54a: U(n) =

N−1

k=0

=

u(kt)e−2π ikn/N t

N−1

N−1

k=0

m=0

 g(mt)h{(k − m)t}t e−2πikn/N t

(9.59)

Changing the order of summation in Equation 9.59, we get U(n) =

N−1

N−1

m=0

k=0

 h{(k − m)t}e

−2π ikn/N

t g(mt)t

(9.60)

Substitution of k − m = α in Equation 9.60 gives U(n) =

N−1

m=0

N−1−m

 h(αt)e

−2π iαn/N

t g(mt)e−2πimn/N t

(9.61)

α=−m

Because h(αt) is periodic and the summation from α = −m to (N − 1 − m) extends over a period, N−1−m

α=−m

h(αt)e−2π iαn/N t =

N−1

h(αt)e−2π iαn/N t

α=0

= H(n)

(9.62)

418

Dynamics of structures

where the last relationship follows from the definition of discrete Fourier transform (Eq. 9.54a). Substitution of Equation 9.62 into Equation 9.61 gives

U(n) =

N−1

H(n)g(mt)e−2π imn/N t = H(n)G(n)

(9.63)

m=0

which proves Equation 9.58. The application of discrete transforms in the evaluation of the response of a linear system can now be summarized as follows. 1

Obtain the discrete transform of a periodic extension of the applied load g(t).

G(n) =

N−1

g(kt)e−2π ikn/N t

(9.64)

k=0

2

Obtain the discrete transform of a periodic extension of the unit impulse function h(t). This will usually require a truncation of h(t) at a time equal to or less than the selected period T0 .

H(n) =

N−1

h(kt)e−2π ikn/N t

(9.65)

k=0

3

Obtain the inverse transform of the product of G(n) and H(n). U(n) = G(n)H(n) u(kt) =

1 2π

N−1

U(n)e2πikn/N 

(9.66a) (9.66b)

n=0

The ability of the discrete response u(kt) to reproduce the continuous response depends on how well the discrete convolution represents the continuous convolution. It is of interest to compare the foregoing expressions for the evaluation of discrete response with the expressions for the response to a periodic load that we derived directly in Section 9.5. On comparing Equations 9.28b and 9.64 and noting that cn T0 is the same as G(n), we find that the two expressions are similar except that the integral over a period in Equation 9.28b is replaced by a summation in Equation 9.64. The complex frequency function H(n) given by Equation 9.65 is a discrete version of H obtained from Equations 9.26 or 9.27. Finally, Equation 9.66b for discrete response is equivalent to Equation 9.28a except that rather than including an infinite number of frequencies, we have restricted the summation to over a finite number of frequencies.

Analysis of response in the frequency domain 419

9.11

COMPARISON OF CONTINUOUS AND DISCRETE FOURIER TRANSFORMS

In application to structural dynamics, the Fourier transform method is used primarily to obtain the convolution of two time functions, one of them being the forcing function and the other the unit impulse response function. As outlined in Section 9.10, the first step in the analysis is computation of the Fourier transforms of the two functions. In most practical applications, the transforms used in the analysis are discrete rather than continuous. A product of the two transforms is obtained next. The inverse transform of this product gives the desired response. It is evident that the evaluation of the transforms represents an intermediate step in the computation and the transforms themselves are of no direct interest. Thus there would be no reason to examine how closely a discrete transform represents a continuous Fourier transform. Occasionally, however, it is not the unit impulse response function, but its continuous Fourier transform, the complex response function, which is specified. In such a situation, the given continuous transform must be converted into a discrete form so that it can be used in forming a product with the discrete transform of the forcing function. The inverse transform of the product gives the required response. It therefore becomes necessary to examine the relationship between the discrete and continuous transforms of a function and to study the adjustments that must be made to the continuous transform so that it resembles the corresponding discrete transform. Consider the rectangular pulse function shown in Figure 9.5a. We first construct a periodic extension of the function with a period T0 = 1.6 s, and then sample the function at N time points within a period, N being equal to 16. The sampled time intervals are given by kt, where k varies from 0 to N − 1 and t = 0.1 s. At a point of discontinuity, the continuous Fourier transform converges to a value that is an average of the values on either side of the discontinuity. Therefore, for a proper comparison of the continuous and the discrete Fourier transform, we define the sampled value at a discontinuity to be equal to such an average. Figure 9.5b indicates the sampled values at k = 0 and k = 4, where the function is discontinuous. We now use Equation 9.54a to obtain the discrete Fourier transform of the sampled function. Figure 9.5c shows the real part of the discrete transform over one period, while Figure 9.5d shows the imaginary part. The horizontal axes in these diagrams represent the frequency scale given by n or nf , where n varies from 0 to N − 1 and f is the frequency increment in hertz. We have f =

 1 = 2π T0

(9.67)

giving  = 1.25π rad/s or f = 0.625 Hz. The discrete Fourier transform is periodic with a period of N = 20π rad/s or Nf = 10 Hz. Figure 9.5c and d show that the real part of the discrete transform is symmetric about n = N/2 while the imaginary part is antisymmetric about that point. We prove below that this is to be expected. For a frequency of m, the discrete Fourier transform is obtained from Equation 9.54a by substituting m for n: G(m) =

N−1

k=0

g(kt)e−2π ikm/N t

(9.68)

420

Dynamics of structures

Figure 9.5 (a) Rectangular pulse function; (b) sampled time function; (c) real part of DFT; (d) imaginary part of DFT.

Analysis of response in the frequency domain 421

A frequency that is symmetric to m about N/2 is l, where l = N/2 + (N/2 − m) =N−m

(9.69)

The Fourier transform at the symmetric frequency is obtained by setting n = N − m in Equation 9.54a: G{(N − m)} =

N−1

g(kt)e−2π ik(N−m)/N t

k=0

=

N−1

g(kt)e2πikm/N t

(9.70)

k=0

It is easily seen that Equation 9.70 is the complex conjugate of Equation 9.68, so that the real parts of the two are identical, while the imaginary parts are equal in magnitude but have opposite signs. It is also of interest to examine the discrete transform value for n = −m. Equation 9.54a gives G{(−m))} =

N−1

g(kt)e2πikm/N

(9.71)

k=0

Equations 9.70 and 9.71 are identical, indicating that the transform values for frequencies higher than N/2 are the same as those for the corresponding negative frequency values. The frequency N/2 is referred to as the folding frequency or the Nyquist frequency and is the highest frequency represented in the transform. The folding frequency should be chosen to be sufficiently large so that the highest frequency of significance in the time function is represented. Since N/2 is equal to π/t, this implies that t should be chosen to be sufficiently small. From a physical point of view this means that if the time function shows rapid variation with time, implying significant high frequency content, it should be sampled more frequently. The continuous Fourier transform of the rectangular pulse function is obtained from Equation 9.32. If t1 denotes the duration of the rectangular pulse,  G() = 

e−it dt

0 t1

= 0

=

t1

 cos t dt − i

t1

sin t dt 0

sin t1 cos t1 − 1 +i  

(9.72)

For purposes of comparison, the real and imaginary parts of the continuous Fourier transform are also plotted in Figure 9.5c and d, respectively.

422

Dynamics of structures

Figure 9.6 Aliasing effect in the Fourier transform of a sampled time function: (a) continuous transform of rectangular pulse; (b) two shifted images of the continuous transform; (c) superposition of (a) and (b).

It is noted that although the magnitude of continuous Fourier transform decreases as the frequency increases, it is never quite zero for all values of frequency higher than any selected value. On the other hand, because the discrete Fourier transform is periodic with a period of N, transform values for frequencies higher than N/2 cannot be represented by the discrete Fourier transform. Within the range of frequencies from 0 to N/2, the continuous and discrete transforms are quite close. The difference between the two becomes more significant as the frequency approaches the folding frequency. This is because of the aliasing effect explained below. Figure 9.6a shows the real part of the continuous Fourier transform of the rectangular pulse function of Figure 9.5a. Let the pulse function be sampled at sample interval t = 0.2 s. The Fourier transform of the sampled waveform is different from that of the continuous version, but can be obtained from that of the latter as follows.

Analysis of response in the frequency domain 423

It can be shown that the Fourier transform of the sampled waveform is periodic with a period equal to 2π/t rad/s or 1/t = 5 Hz. To obtain the transform of sampled time function, we first construct a series of frequency functions of the same shape as that of the continuous transform but shifted on the frequency axis by 2π m/t rad/s, where m is an integer that may range from −∞ to +∞. Superposition of these frequency functions gives the desired transform. Figure 9.6b shows two versions of the continuous transform function, one shifted by +2π/t and the other by −2π/t. A superposition of Figure 9.6a and b is shown in Figure 9.6c. The resulting transform is different from the continuous transform because of the overlapping effect. It is obvious that such overlapping will always occur unless the continuous transform is zero for frequency values greater than π/t. Frequency functions that have zero values outside a certain range of frequencies are called bandlimited. For such transform functions, overlapping, also known as aliasing, can be avoided if t is chosen such that π/t, the folding frequency, is higher than the highest frequency in the transform. This leads to the important conclusion that if the transform of the sampled time function is required to closely approximate that of the continuous time function, the sampling intervals should be chosen sufficiently small so as to avoid aliasing. For transform functions that are not bandlimited, aliasing will occur whenever a time function is sampled. However, if the transform value decreases with frequency, the aliasing effect can be reduced by making t sufficiently small. It will also be noted that for transform functions that decrease in value with frequency, the effect of aliasing is most significant near the folding frequency. We have seen that sampling in the time domain leads to a transform function that is periodic, the period being equal to 2π/t rad/s. In a similar manner, sampling of the frequency function will lead to a time-domain function that is periodic with a period given by T0 = 2π/. Also, just as sampling in the time domain resulted in aliasing in the frequency domain, sampling in the frequency domain will result in aliasing in the time domain, unless the function is time limited (is zero outside a certain time range), and  is sufficiently small to avoid time function overlap. In the example under consideration the value of T0 = 1.6 s is sufficiently large so that time-domain aliasing does not occur. It is of interest to note the effect of choosing T0 = 0.8, which gives  = 2.5π rad/s or 1.25 Hz. The discrete Fourier transform values will be those in Figure 9.5c corresponding to n = 0, 2, 4, 6, and 8. The first of these values is equal to 0.4; all others are zero. When these results are compared with continuous Fourier transform values, it is obvious that the sampling interval for the frequency function is so coarse that the function is not adequately represented. In summary, if the discrete transform has to closely approximate the continuous transform t should be significantly small so that the time function is adequately represented and T0 should be sufficiently large so that the frequency function is adequately represented. Example 9.2 Compare the continuous and discrete Fourier transforms of a unit impulse function. The undamped natural frequency of the system is 2π rad/s, the sampling interval t = 0.1 s and m = 0.25 lb · s2 /in. Assume that the damping ratio is zero.

424

Dynamics of structures

Solution The unit impulse function h(t) given by  1 sin ωt t > 0 h(t) = mω 0 t0 t 0 h(t) = mω (9.74) 0 ti

(11.24)

and a˜ ij = a˜ ji = (ajj − aii )sin α cos α + aij (cos2 α − sin2 α)

(11.25)

By equating the off-diagonal term to zero, we obtain tan 2α = α=

2aij aii − ajj

aii = ajj

(11.26a)

π 4

aii = ajj

(11.26b)

Each transformation affects all elements in rows and columns i and j. Apparently, therefore, an off-diagonal element reduced to zero during a transformation will become nonzero during subsequent transformations. The process is, in fact, iterative. With repeated iterations, the off-diagonal elements reduce in magnitude until they are negligible in comparison to the diagonal values. After a sufficient number of iterations, the transformed matrix becomes diagonal. The eigenvalues of the diagonal matrix are equal to the elements on the diagonal. Since the transformations do not alter the eigenvalues, the eigenvalues of the diagonal matrix are also those of the original. The eigenvectors of the diagonal matrix are unit vectors, hence Q = I, the identity matrix. The eigenvectors of the original matrix are obtained by using the relationship Q = N1 N2 · · · Nm Q

(11.27)

In the standard Jacobi diagonalization procedure, the off-diagonal elements with the largest absolute value are chosen for elimination at each step. In a computer implementation, the process of finding the largest element is likely to be quite time consuming, adding to the cost of computation. A variation of the process in which the algorithm does not involve such a search is therefore generally preferred. One possible alternative is to eliminate off-diagonal elements in a systematic order, for example, (i, j) = (1, 2), (1, 3), . . . , (1, N) followed by (2, 3), (2, 4), . . . , (2, N), and so on. This process, called serial Jacobi, still converges to the right solution, although the convergence may be slower.

Numerical solution of the eigenproblem

531

In another variation, a threshold value is established, and although the elements are eliminated in a serial order, elimination is skipped if the modulus of the off-diagonal element under consideration lies below the threshold. After a complete pass through the matrix, the threshold value is lowered and the process repeated until the transformed matrix is nearly diagonal. Since the objective is to reduce the magnitude of off-diagonal elements in comparison to those of the diagonal, comparison of element (i, j) with a threshold is usually accomplished by using the following inequality 

(aij )2 aii ajj

1/2

< 10−s

(11.28)

where s is a measure of the desired convergence. For its solution by the standard Jacobi diagonalization process, the eigenvalue problem of Equation 10.7 must first be reduced to the standard symmetric form given by Equation 10.5. However, a variation of the method, called the generalized Jacobi method, can be applied directly to Equation 10.7. The procedure consists of repeated simultaneous transformations of K and M matrices until they are both reduced to a diagonal form. Mathematically, the transformation can be expressed as NT KNN−1 q = λNT MNN−1 q

(11.29a)

˜ q˜ = λM ˜ q˜ K

(11.29b)

or

˜ = NT KN, M ˜ = NT MN, and q = Nq. ˜ in which K The transformation matrix N is given by



1 .  i 0 N=  .  j 0 0

i . 0 . . . 1 . . . α . 0

j . 0 . . . γ . . . 1 . 0

 . 0 . .  . 0   . .  . 0 . 1

(11.30)

On carrying out the transformations and equating the off-diagonal elements k˜ ij and m ˜ ij of the transformed stiffness and mass matrices to zero, we get kii γ + kjj α + kij (1 + αγ ) = 0 mii γ + mjj α + mij (1 + αγ ) = 0

(11.31)

532

Dynamics of structures

Solution of Equation 11.31 gives a = kjj mij − kij mjj b = kii mjj − kjj mii c = kii mij − kij mii  1 [b + sign(b) b2 + 4ac] 2a c α=− b a γ = −α c α=

a = 0

(11.32)

a=0

where kij and mij represent elements of K and M, respectively, and i < j. If a and b are simultaneously zero, α is taken as zero and γ = −kij /kii . In the expression for α, the square-root term is taken with the sign of b to avoid having to take the difference of two quantities that may be of similar magnitude. This improves the numerical stability of the method. Use of the generalized Jacobi method avoids the need for transforming the linear eigenvalue problem to a standard form which may involve a substantial amount of computations and, in addition, may introduce numerical inaccuracies when the matrices are ill conditioned. In particular, when the off-diagonal elements of the matrices are small, conversion to a standard form will give a matrix with a combination of very large and very small numbers. Application of the standard Jacobi technique is, in such cases, likely to result in numerical inaccuracies. Finally, when the stiffness and mass matrices are banded so that the off-diagonal element are already zero, use of the generalized Jacobi method for reduction of the matrices to a diagonal form is very efficient. Conversion to a standard form, on the other hand, destroys the bandedness of the matrices so that no advantage can be taken of their sparsity. Example 11.1 Use Jacobi’s method to determine the eigenvalue of the symmetric matrix A given below.   2640 230 −1330 0  230 770 0 0  A= −1330 0 2640 −230 0 0 −230 770

Solution The standard Jacobi transformation given by Equations 11.22 and 11.26 is used. In forming the transformation matrix (Eq. 11.22), we need sin α and cos α. These are obtained directly from Equation 11.26 by using the following expressions aii − ajj 2R aii − ajj 2 sin α = 0.5 − 2R aij sin α cos α = R cos2 α = 0.5 +

(a)

Numerical solution of the eigenproblem

533

in which 1 21/2 R = (aii − ajj )2 + 4a2ij Function sin α is chosen as positive so that cos α takes the sign of aij . At each iteration step the off-diagonal element with the largest absolute value is reduced to zero. The iteration steps are listed below. Iteration no.

Row i

Col j

cos α

Transformed matrix NT AN

sin α 

1

1

3

−0.7071

0.7071

3970.00 −162.63

−162.63 770.00 −162.63    0 −162.63 1310.00 −162.63 

3978.20

2

1

4

−0.99872

0.05063

   

 3

2

3

−0.26777

0.96348

0 −162.63

.. .

.. .

.. .

162.63

770.00

162.40

8.20

0

8.20   8.20 −162.63 1310.00 −162.40 770.00 −162.63

0

8.20 −162.40

761.80

3978.20 −35.56 −158.70

0

 −35.56 1355.20   −158.70 0

3

−0.999994

0.011111



0 −158.70   724.80 35.56 35.56

761.80

.. .

.. . 3986.40

2



162.4

 11

0   162.63

0

0 −158.70 .. .



   

0.06

0

0



0

0   0 753.55 −0.06

0

0 −0.06 685.32

0.06 1394.70

0

The largest off-diagonal element in the original matrix is in row 1 and column 3, its absolute magnitude being 1330. In the first iteration step element a13 and a31 are reduced to zero. In the second iteration step, a14 and a41 are reduced to zero. It should be noted that following the latter transformation, element a13 and a31 are restored to nonzero values. At the end of 11 iteration cycles, the off-diagonal elements are all zero up to the first decimal digit. The diagonal elements at this stage give the desired eigenvalues. Thus λ1 = 685.32 λ2 = 753.55 λ3 = 1394.7 λ4 = 3986.4

534

Dynamics of structures

If desired, the eigenvectors can be obtained from Equation 11.27. Substitution of N1 , ˜ = I gives N2 , . . . , N11 and Q 

−0.2443  0.6636 Q= −0.2443 −0.6636

−0.0504 0.7053 0.0504 0.7053

0.6636 0.2443 0.6636 −0.2443

 0.7053 0.0504  −0.7053 0.0504

11.4.2 Householder’s transformation Householder’s method reduces a symmetric matrix to tridiagonal form by the application of a series of orthogonal transformation. The eigenvalues and eigenvectors of the tridiagonal matrix are then computed by either the LR or the QR method, to be described later. The eigenvalues so computed are the same as those of the original matrix, while the eigenvectors of the tridiagonal matrix are related to those of the original by a relationship of the form of Equation 11.21b. Since Householder’s method can be applied to a symmetric matrix only, the linearized eigenvalue problem must first be converted to a standard symmetric form. To illustrate the method, consider Figure 11.2, which shows the transformed matrix A after reduction has been completed to row k − 1. At this stage, the matrix is tridiagonal from row 1 to k − 1. In the kth transformation, we wish to reduce to zero

Figure 11.2 Householder’s transformation: (a) matrix A; (b) transformation matrix N.

Numerical solution of the eigenproblem

535

all nonzero elements in row k as well as column k except element a¯ k,k−1 , a¯ k,k , a¯ k,k+1 , a¯ k−1,k , and a¯ k+1,k , where the bar on a indicates that the current value of the element is taken. The elements to be reduced to zero are enclosed in boxes in Figure 11.2. The kth transformation is given by Ak+1 = NkT Ak Nk

(11.33)

The transformation matrix Nk is shown in a partitioned form in Figure 11.2b. It consists of submatrices I (the identity matrix) and R and two null matrices. The submatrix R is selected to be of the form R = I − 2wwT

(11.34)

where w is a column vector of size N − k with Euclidean norm equal to unity, so that wT w = 1

(11.35)

By taking the transpose of Equation 11.34, it is seen that RT = R. It follows that NT = N. Thus both R and N are symmetric. Further, RRT = (I − 2wwT )(I − 2wwT ) = I − 4wwT + 4wwT wwT

(11.36)

Substitution of Equation 11.35 into 11.36 gives RRT = I

(11.37)

By using the relationship in Equation 11.37, it can be proved that NNT = I. This implies that the transformation specified by Equation 11.33 is orthogonal. Expansion of the basic transformation Equation 11.33 using the partitioned forms of A and N shown in Figure 11.2 gives   B 0  cT R   (11.38) Ak+1 =    0 Rc RDR We note from Equation 11.38 that the transformation leaves the submatrix B unaltered. Thus the computations in the kth step do not change that portion of A which has already been tridiagonalized. To determined w and hence R, we note that the matrix product Rc should give a vector all of whose elements except the first are zero. Thus Rc = g

(11.39a)

or 

 I − 2wwT c = g

(11.39b)

536

Dynamics of structures

where gT = [r

0

cT = [¯ak,k+1

···

0

a¯ k,k+2

0], r is as yet undermined and ···

a¯ k,N ]

(11.40)

From Equation 11.39a we have gT g = cT RT Rc

(11.41a)

r2 = cT c

(11.41b)

or

Now premultiplying both sides of Equation 11.39b by cT and denoting the scalar quantity wT c = cT w by h, we get cT c − 2h2 = cT g

(11.42)

Noting that cT g = a¯ k,k+1 r and using Equation 11.41b, Equation 11.42 reduces to 2h2 = r2 − a¯ k,k+1 r

(11.43)

Equations 11.41b and 11.43 provide the values of r and h. The sign of r is chosen so that the product a¯ k,k+1 r in Equation 11.43 is negative. Now using Equation 11.39b and that wT c = h, 2hw = c − g ≡ v

(11.44)

Finally, Equation 11.34 leads to the transformation matrix R R=I−

1 vvT 2h2

(11.45)

in which v = c − g and 2h2 is given by Equation 11.43. The Householder transformation begins with A1 = A and the vector c given by cT = [a12

a13

···

a1N ]

(11.46)

It should be noted that at each step the transformed matrix Ak remains symmetric and hence only half of the matrix need be kept in the computer storage. However, in the case of a banded matrix, the bandwidth of the unreduced part during an intermediate steps may be larger than that of the original matrix.

Numerical solution of the eigenproblem

537

Example 11.2 Tridiagonalize the matrix in Example 11.1 by using Householder’s method.

Solution In the first step of transformation, elements a31 , a41 and a13 , a14 are reduced to zero. The steps involved in the computations are as follows  230.0 c1 = −1330.0 0.0 

(a)

r21 = c1T c1 = 1,821,800 2h21

=

r21

(b)

− a12 r1

= 1,821,800 − 230 × (−1349.74)

(c)

= 2,132,240   −1349.74 0  g1 =  0

(d)

v1 = c1 − g1   1579.74 = −1330.00 0  1.17040 −0.98537 v1 v1T 0.82960 = −0.98537 2 2h1 0 0 R1 = I − 

1 v1 v1T 2h21

−0.17040 =  0.98537 0

0.98537 0.17040 0

(e)  0 0 0

(f)

 0 0 1

(g)

The transformed matrix obtained from Equation 11.38 becomes 

2640.0 −1349.7  A2 =  0 0 In the second step,  314.0 −226.6

 c2 =

r22 = 149,937 2h22 = 271,547

−1349.7 2585.7 314.0 −226.6

0 314.0 824.3 −39.2

 0 −226.6  −39.2 770.0

538

Dynamics of structures v2 = R2 =

 701.23

−226.64 −0.81085 0.58526

 0.58526 0.81085

At the end of second step, the transformed matrix, which is now tridiagonal, is given by 

2640.0

 −1349.7 A3 =   0  0

−1349.7 2585.7

0 −387.24

−387.24

842.90

0

−13.42

0



   −13.43  751.40 0

A major portion of the computations involved in Householder’s method are those required for the evaluation of matrix product RDR. For general matrices, the triple matrix product will need 2m3 multiplications, where m is the order of each of the matrices. However, because of the special nature of R, the number of multiplications can be reduced substantially by using the following steps in the calculations s = Dw pT = 2sT β = pT w

(11.47)

q = p − 2βw RDR = D − wpT − qwT The validity of the series of steps in Equation 11.47 can be verified by direct substitution. The number of multiplications is now 3m2 + 3m. For large m this is substantially smaller than 2m3 .

11.4.3 QR transformation The QR transformation, which derives its name from the notations used in the transformation, can be applied to any symmetric matrix to reduce the latter to a diagonal form. However, it is usually more efficient first to reduce the matrix to a tridiagonal form by using Householder’s method and then to apply the QR transformation to reduce the tridiagonal matrix to a diagonal form. As described earlier, the eigenvalues and eigenvectors of the diagonal matrix are obtained by inspection. In the following paragraph we describe the application of QR method to the reduction of a tridiagonal matrix. Extension to the case of a more general symmetric matrix is straightforward. The QR transformation is essentially iterative in nature. In the kth iteration step, matrix Ak is transformed to Ak+1 . This is done in two steps. Matrix Ak is first

Numerical solution of the eigenproblem

539

transformed to an upper triangular form. This is achieved by premultiplying Ak by T . Matrix Ni is given by a series of orthogonal matrices N1T , N2T , . . . , NiT , . . . , NN−1 

i

1 0 . 0  . . . .  T  i 0 0 . cos α Ni =   i + 1  0 0 . −sin α  . . . .  0 0 . 0

i+1 0 . sin α cos α . 0

 . 0  . .  . 0   . 0  . .  . 1

(11.48)

With α chosen appropriately, premultiplication by NiT reduces the subdiagonal term ai+1,i in the current version of A to zero. Also, the subdiagonal terms already reduced to zero during earlier operations are unaltered. For ease of reference, we (i) (i) T , . . . , N2T N1T Ak by Ak . The subdiagonal terms up to row i in matrix Ak , represent Ni−1 (i) have already been reduced to zero. On forming the product NiT Ak and equating the term ai+1,i to zero, we get −¯aii sin α + a¯ i+1,i cos α = 0

(11.49)

so that tan α =

a¯ i+1,i a¯ ii

(11.50) (i)

where the bar on an element a indicates that its value from Ak has been used. T On completing the full series of transformations using N1T , N2T , . . . , NN−1 , we get T · · · N2T N1T Ak = Rk NN−1

(11.51)

QkT Ak = Rk

(11.52)

or

where Rk is an upper triangular matrix and T · · · N2T N1T QkT = NN−1

(11.53)

T are orthogonal, QkT is also orthogonal, Since each of the matrices N1T , N2T , . . . , NN−1 −1 T that is, Qk = Qk . The transformation to Ak+1 is now completed by using the expression

Ak+1 = Rk Qk

(11.54)

where Qk = N1 , N2 , . . . , NN−1

(11.55)

540

Dynamics of structures

Substitution of Equation 11.52 into Equation 11.54 gives Ak+1 = QkT Ak Qk

(11.56)

It is clear from Equation 11.56 that the transformation from Ak to Ak+1 is orthogonal under which the eigenvalues remain unchanged. The matrix product on the left-hand side of Equation 11.51 strips all subdiagonal terms in Ak so that the product matrix Rk is an upper triangular matrix. However, when the kth iteration is completed by carrying out the matrix operations indicated in Equation 11.54, the subdiagonal elements will, in general, be restored to nonzero values, but their magnitudes would have reduced. Eventually, after sufficient number of iterations, matrix A would become almost diagonal, the elements on the diagonal being equal to the eigenvalues of the original tridiagonal matrix. If required, the eigenvectors of the tridiagonal matrix can be obtained by using an expression similar to Equation 11.21b: T = Q1 Q2 · · · Qk I

(11.57)

In Equation 11.57, I, the identity matrix, is the matrix of the eigenvectors of a diagonal matrix, T is the matrix of eigenvectors of the tridiagonal matrix, Q1 , Q2 , . . ., Qk are the orthogonal transformation matrices, and it is assumed that convergence has been achieved after k iterations. In practice, it is more efficient to evaluate the eigenvectors by using the inverse iteration method described later. In that case only those eigenvectors that are of interest need be evaluated. When a combination of the Householder reduction, QR transformation and inverse iteration is used for the evaluation of eigenvalues and eigenvectors, the method is commonly referred to as the HQRI method. In the QR iteration procedure described above, eigenvalues are obtained in increasing order of magnitude. The lowest eigenvalue, λ1 , is obtained first and occupies the position a¯ NN . When this element does not change significantly with successive iterations, a¯ NN has converged to λ1 . At this stage, if desired, the matrix can be deflated by removing the Nth row and Nth column and the iterations continued on the resulting N − 1 by N − 1 matrix until the next eigenvalue λ2 is obtained. This process is continued, the ith eigenvalue being obtained by iterations on a deflated matrix of size N − i + 1. Assuming that eigenvalues λ1 , λ2 , . . ., λi−1 have already been obtained, convergence to λi depends on the ratio |λi /λi+1 |; the smaller this ratio, the faster is the convergence. This suggests that convergence can be improved by using a modified procedure called QR iteration with shifted origin. The procedure is based on the fact ˜ = A − µI, where µ is any scalar value, are related that the eigenvalues λ˜ of the matrix A to the eigenvalues of A by the expression λ˜ j = λj − µ

(11.58)

Equation 11.58 can easily be proved by noting that ˜ = Aq − µq Aq = (λ − µ)q

(11.59)

Numerical solution of the eigenproblem

541

In the kth iteration of the QR method for obtaining λi , an estimate, even though approximate, has already been obtained for the desired eigenvalue. The estimated value denoted by µk may be taken as equal to the element aii of the current version of A. The next transformation is now carried out on matrix (Ak − µk I), so that ˜k ˜ T (Ak − µk I) = R (11.60) Q k

˜ k+1 = R ˜k ˜ kQ A

(11.61)

The transferred matrix is then restored by removing the shift. Thus ˜ k+1 + µk I Ak+1 = A

(11.62)

A new estimate is now obtained for λi . If convergence has not been achieved, the iteration process (Eqs. 11.60, 11.61, and 11.62) is repeated. It is evident that if the shift is close to the value of λi , the ratio |λ˜ i /λ˜ i+1 | will be small and the convergence will be speeded up significantly. Example 11.3 Reduce the following tridiagonal matrix to a diagonal form by using QR iteration   3 −2 0 8 −4 A = −2 0 −4 5

Solution The first transformation matrix Q1 in the series of orthogonal transformation to be performed on A is itself the product of two other matrices and is given by Q1 = N1 N2

(a) N1T A

the subdiagonal element a21 is zero. Matrix N2 eliminates in which N1 is selected so that in element a32 without affecting a21 , so that in the product N2T N1T A both a21 and a32 are zero. Computations for obtaining N1 and N2 are tabulated below. In each case, matrix N is given by Equation 11.48 with α determined from Equation 11.50. Using the later equation, the following expressions are obtained for cos α and sin α a¯ ii cos α = * 2 a¯ ii + a¯ 2i+1,i a¯ i+1,i sin α = * 2 a¯ ii + a¯ 2i+1,i

i

cos α

(b)

Transformed matrix, A(i+1) = NiT A(i)

sin α 

1

0.83205

−0.55470

2

0.81111

−0.58490

3.606 0 0  3.606 0 0

−6.102 5.547 −4.000 −6.102 6.839 0

 2.219 −3.328 5.000  2.219 −5.624 2.109

542

Dynamics of structures

Note that in the above, A(1) = A and the matrix R1 = A(3) obtained at the end of second step is an upper triangular matrix. The first transformation matrix Q1 is now given by Q1 = N1 N2  0.83205  = −0.55470 0

0.44992 0.67468 −0.58490

 0.32444  0.48666 0.81111

The first transformation is completed by carrying out the matrix multiplications in A2 = Q1T AQ1 = R1 Q1

(c)

which gives 

6.385  A2 = −3.794 0

 0  −1.235 1.711

−3.794 7.905 −1.234

At the end of eight transformations we will get 

11.1110  A9 = −0.0033 0

−0.0033 3.5820 −0.0008

 0  −0.0008 1.3060

At this stage, we have nearly converged to the lowest eigenvalue of 1.306, and if accuracy to about three decimal digits is considered adequate, the matrix can be deflated by omitting the last row and the last column. In this case, however, we will continue our iteration on the complete matrix. Four figure accuracy is achieved at the end of 12 transformation, giving 

A13

11.111  = 0 0

0 3.582 0

 0  0  1.306

The eigenvalues of the given matrix are thus λ1 = 1.306 λ2 = 3.582 λ3 = 11.111

11.5

ITERATION METHODS

Iteration methods are ideally suited for eigenvalue problems of large size with sparse or banded matrices, particularly when only some of the eigenvalues, either the lowest few or the highest few, are required. The following iteration methods have been effectively

Numerical solution of the eigenproblem

543

applied to practical solutions of eigenvalue problems in the vibration of engineering structures: 1 2 3 4

Vector iteration Vector iteration with shift Subspace iteration Lanczos method

A brief description of each of the methods is provided in the following paragraphs.

11.5.1 Vector iteration Direct vector iteration or simply vector iteration is also known as the power method. It allows successive evaluation of as many eigenvalues and eigenvectors as required, starting with the eigenvalue with the largest absolute value. Considering the standard eigenvalue problem assume that the eigenvalues have been arranged so that |λ1 | < |λ2 | < |λ3 | < · · · < |λN |. Begin with an arbitrary vector u and let it be expressed in the form of Equation 11.9. Premultiplication of Equation 11.9 by A gives Au = c1 Aq1 + c2 Aq2 + · · · + cN AqN = c 1 λ1 q 1 + c 2 λ2 q 2 + · · · + c N λN q N

(11.63)

It is easily seen that after k multiplications by A Equation 11.9 will give Ak u = c1 λk1 q1 + c2 λk2 q2 + · · · + cN λkN qN

k λ1 k λ2 k = λ N c1 q1 + c 2 q2 + · · · λN λN 
 λN−1 k + cN−1 qN−1 + cN qN λN

(11.64)

Now because |λi /λN | < 1 for i = 1, 2, 3, . . . , N − 1, in the limit as k becomes large uk = Ak u  cN λkN qN

(11.65a)

uk+1 = Ak+1 u  cN λk+1 N qN

(11.65b)

and

The rate of convergence depends on the ratio λN−1 /λN ; the smaller this ratio, the faster the rate of convergence. The product vector obtained from either Equation 11.65a or 11.65b is proportional to the eigenvector that corresponds to the eigenvalue with the largest magnitude.

544

Dynamics of structures

Also, the ratio of the value of an element in the (k + 1)th iteration to the value of the same element in the kth iteration approaches λN as k becomes large. Rather than using the ratio of the two successive values of an arbitrarily selected element of the product vector, the following alternative expression may be used to obtain a better estimate of the eigenvalue after k + 1 iterations:

λN ≈ ρk+1 =

T uk+1 uk+1 T uk+1 uk

(11.66)

Equation 11.66 can easily be shown to be true by substituting for uk+1 and uk from Equations 11.65b and 11.65a, respectively. In practice, because the magnitude of an eigenvector is arbitrary, it is convenient to normalize the product vector at the end of each iteration. In fact, with each multiplication by A, the elements of the product vector grow or decrease in magnitude, and this growth or decrease may cause numerical problems in computations. Scaling of the vector after each iteration keeps element values within reasonable bounds. Scaling can be achieved either through a division by the largest element of the vector or by normalizing the vector so that its Euclidean norm is unity. When the latter method is adopted, the procedure is represented by the following steps u¯ k+1 = Auk ρk+1 = uk+1 =

(11.67)

T u¯ k+1 u¯ k+1

(11.68)

T u¯ k+1 uk

u¯ k+1 T (u¯ k+1 u¯ k+1 )1/2

(11.69)

Equation 11.68 is a Rayleigh quotient estimate of the eigenvalue, and Equation 11.69 represents normalization of the vector u¯ k+1 so that the normalized vector uk+1 has a Euclidean norm of unity. Example 11.4 Using direct vector iteration, determine the eigenvalue with the largest absolute magnitude and the corresponding eigenvector of matrix A given by 

−64 A = −136 72

48 102 −54

 36 76 −40

Solution We use the iteration procedure defined by Equations 11.67, 11.68, and 11.69 using for our initial trial vector a vector all of whose elements are equal to unity. The computations are tabulated as follows.

Numerical solution of the eigenproblem

k

uk =

u¯ * k u¯ kT u¯ k

T u¯ k+1 uk

 20  42 −22

2648.0

40.0

66.2

 29.69  62.88 −33.19

5936.8

48.68

77.05

 0.3853  0.8161 −0.4307

  −0.9916 −1.8918 0.9002

5.373

−2.314

−2.322

 −0.4278 −0.8162 0.3884



 2.184  4.447 −2.263

29.67

−5.443

−5.450

 0.4010  0.8165 −0.4155



 −1.430  2.831 1.401

12.02

−3.467

−3.467

 −0.4124 −0.8165 0.4041



 1.748  3.515 −1.766

18.53

−4.305

−4.305

 0.4061  0.8166 −0.4103



 −1.564 −3.119 1.555

14.59

−3.820

−3.820

 −0.4095 −0.8164 0.4070



 1.673  3.351 −1.678

16.85

−4.104

−4.104

 0.4076  0.8165 −0.4089



 −1.615 −3.227 1.612

15.62

−3.952

−3.952

 −0.4086 −0.8165 0.4079



 1.643  3.287 −1.644

16.21

−4.026

−4.026

 0.4081  0.8165 −0.4084



 −1.629 −3.257 1.628

15.912

−3.989

−3.989

1

 0.3887 0.8162 0.4275 

2

 3

 4

 5

 6

 7

 8

 9

 10

 11

 −0.4084 −0.8165 0.4081 

12

ρ=

T u u¯ k+1 ¯ k+1

T u¯ k+1 u¯ k+1

  1 1 1

T u¯ k+1 = Auk

545

T u u¯ k+1 k





546

Dynamics of structures

At this stage convergence may be assumed to have been achieved. The eigenvalue is −3.99 and the corresponding eigenvector is given by   −0.4084 q1 = −0.8165 0.4081 For comparison the exact eigenvalue is λ1 = −4 and the exact eigenvector is   −0.4082 q1 = −0.8165 0.4082

11.5.2 Inverse vector iteration The vector iteration method can be applied to the linearized eigenvalue problem by reducing to a standard form given by either Equation 11.12b or Equation 11.13b. When applied to the form of Equation 11.12b, iterations will converge to the most dominant value of λ, that is, to the highest frequency ω. When applied to Equation 11.13b, convergence will be to the largest value of γ , that is, to the lowest value of λ or ω. In practical vibration problems, the lower frequencies and mode shapes are of greater interest, and therefore the form given by Equation 11.13b is preferred. Iteration on Equation 11.13b, which leads to the lowest value of ω, is usually referred to as inverse vector iteration. Inverse iteration involves repeated multiplication by matrix D, usually referred to as the dynamic matrix. In all other respects, the method is similar to the one described in Section 11.5.1. The algorithm can thus be summarized as u¯ k+1 = Duk

(11.70)

ρk+1 =

T uk u¯ k+1 T u¯ k+1 u¯ k+1

(11.71)

uk+1 =

u¯ k+1 T (u¯ k+1 u¯ k+1 )1/2

(11.72)

Equation 11.71, which is the inverse of Equation 11.68, gives the reciprocal of the eigenvalue, so that ρk+1 = 1/γ = λ. For the purpose of numerical computations it usually is more efficient to organize the inverse iteration on a linearized eigenvalue problem as follows: Ku¯ k+1 = xk x¯ k+1 = Mu¯ k+1

(11.73) (11.74)

ρk+1 =

T xk u¯ k+1 T u¯ k+1 x¯ k+1

(11.75)

xk+1 =

Mu¯ k+1 x¯ k+1 = T (u¯ k+1 Mu¯ k+1 )1/2 x¯ k+1 )1/2 (u¯ k+1

(11.76)

Numerical solution of the eigenproblem

547

Iteration is commenced with x = Mu, where u is an arbitrary trial vector. Equation 11.73 is solved for u¯ k+1 , the solution being equivalent to multiplying the current version of the trial vector by D = K−1 M. Equation 11.75 is a Rayleigh quotient estimate of the eigenvalue. Equation 11.76 normalizes the current value of the trial vector T Muk+1 = 1. When k is sufficiently large, ρk+1 converges so that xk+1 = Muk+1 and uk+1 2 to λ1 = ω1 and uk+1 converges to φ1 , so that λ1 = ρk+1 φ1 =

(11.77a)

u¯ k+1 T (u¯ k+1 x¯ k+1 )1/2

(11.77b)

Convergence can be measured by comparing two successive values of λ. Thus (k+1)

|λ1

(k)

− λ1 |

(k+1)

2, there is a possibility of jumping over one or more roots. Fortunately, this can be detected by using the Sturm sequence property so that if necessary the iteration can be retraced. A major computation effort is required in evaluating the determinant for a given value λ = µ. The evaluation is carried out by obtaining an LDLT factorization of matrix (K − µM). The determinant of the matrix is then obtained by taking the product of the diagonal elements of D. According to the Sturm sequence property, the number of negative elements in D is equal to the number of eigenvalues less than µ. Skipped eigenvalues, if any, can thus be detected by counting the number of negative elements in D. To begin the iteration for λ1 , two estimates, µ1 and µ2 , that are both smaller than λ1 are required. The first estimate, µ1 , can be taken as zero. Because all eigenvalues are positive, µ1 = 0 is obviously less than or equal to λ1 . To obtain the second estimate, an inverse integration process without shift is used. The Rayleigh quotient estimate of λ1 obtained after a few iterations is then used to select a value for µ2 . The determinant search method can also be employed to converge to higher roots. An additional complication arises this time. For the first root, if the iteration began with two values of µ, both less than λ1 , it should converge toward λ1 . However, for a higher root this is not guaranteed. For example, iteration with points A and B in Figure 11.5a

Figure 11.5 Determinant search method of determining eigenvalues: (a) determinant search for search for λ1 ; (b) determinant search for λ2 .

Numerical solution of the eigenproblem

573

will converge back to λ1 . To overcome this difficulty, a deflated polynomial pj (µ) is used instead of p(µ), where pj (µ) =

p(µ) j i=1 (µ

(11.150)

− λi )

and λi (i = 1 to j) are the eigenvalues already determined. Iteration with pj (µ) will converge to the (j + 1)st eigenvalue, λj+1 . The deflated polynomial for the second root as shown in Figure 11.5b. In practice, the determinant search method is used to obtain only a close estimate of the eigenvalue desired. More accurate estimates of the eigenvalue and the corresponding eigenvector are then obtained by using inverse vector iteration with shifts is described in Section 11.5.3. It is apparent from the foregoing discussion that determinant search is most effective when used to locate a shift close to the eigenvalue desired. Thus each eigenpair is calculated independently of others and the precision obtained in determination of previous eigenvalues has little bearing on the accuracy of the eigenpair currently being calculated. Also, since inverse iteration is used with shifts, orthogonalization of the vector need not be carried out the respect to all previous eigenvectors – just a few. Example 11.8 For a system with the stiffness and mass matrices given below, obtain the eigenvalues and eigenvectors using a combination of determinant search method and inverse iteration with shift. 

3 K = −2 0

−2 8 −4

 0 −4 5





2

M=

4

 2

Solution Since the eigenvalues of a stable system are either equal to zero or positive, we begin the determinant search for the first eigenvalue with two initial estimates, µk−2 = −1 and µk−1 = 0. Both of these values must be less than the true eigenvalue. A new estimate for µ is obtained by using Equation 11.149 with η = 1. The computations are tabulated as follows.

µk = µk−1 −

µk−1 − µk−2 pk−1 pk−1 − pk−2

k

µk−2

pk−2

µk−1

pk−1

1

−1

312

0.0

52

0.2

2

0

52

0.2

26.1

0.4

7.1

3

0.2

26.1

0.4

7.1

0.475

1.65

pk 26.1

It is evident that we are approaching the zero of the polynomial p. At this stage we use inverse iteration with shift to obtain a more accurate estimate of the eigenvalue as well as the corresponding eigenvector.

574

Dynamics of structures With a shift of µ = 0.475, ˆ = [K − µM] K  2.05 −2.00 6.10 = −2.00 0 −4.00

 0 −4.00 4.05

The new eigenvalue problem is ˆ = δMq Kq Inverse iteration is carried out by using Equations 11.73 through 11.76. The initial trial vector is taken as     2 1 y1 = M1 = 4 2 1 k

xk   2 4 2

1

 0.707 1.414 0.707 

2

u¯ k+1   40 40 40   14.14 14.14 14.14

x¯ k+1   80 160 80   28.28 56.57 28.28

T u¯ k+1 xk

T u¯ k+1 x¯ k+1

320

12,800

39.98

1,599.7

xk+1

ρk+1

 0.707 1.414 0.707 

0.025

0.025

Convergence has been achieved at this stage. We therefore have δ = ρ3 = 0.025 λ1 = µ + δ = 0.475 + 0.025 = 0.50 u¯ 3 φ1 =  1/2 T u¯ 3 x¯ 3   0.3535 = 0.3535 0.3535 Determinant search for the second eigenvalue is carried out on a deflated polynomial given by p(2) =

p µ − 0.5

To carry out the iterations, we need two estimates for the next eigenvalues. We will take these estimates as 0.6 and 1.0, respectively. It must be ensured that in doing so we have not skipped the second eigenvalue; that is, none of the two estimates is larger than the second eigenvalue. This could be verified by using a Sturm sequence check. If LDLT decomposition of [K − µM] shows two negatives on D, it is evident that the value of µ selected is larger than the

Numerical solution of the eigenproblem

575

first two eigenvalues and that a lower value should be selected for µ. The computations for the second eigenvalue are tabulated as follows. k

µk−2

p(2) k−2

µk−1

p(2) k−1

µk

1

0.6

−56

1

−32

1.53

−6.68

1

1.0

−32

1.53

1.67

−0.21

−6.68

p(2) k

As this stage we use inverse iteration with a shift of 1.67 to determine the second eigenvalue and the second eigenvector. If the shift point is closer to the second eigenvalue than to the first one, iteration should converge to the second eigenvalue. However, when eigenvalues are coincident or lie close together, the eigenvectors already determined must be swept from the trial vector. In practical applications, it is reasonable to sweep off, say, the six previous eigenvectors. For the present case, for simplicity, we will assume that it is not necessary to sweep off the previous eigenvector. The new stiffness matrix is   2 3 −2 0 ˆ = −2 8 −4 − 1.67 K 0 −4 5   −0.34 −2 0 1.32 −4  = −2 0 −4 1.66







4 2

The computations involved in the inverse iteration are tabulated below. We take care that the initial trial vector is not of the form u1T = [1 1 1], because that being precisely equal to the first mode shape, the iteration is likely to converge to the first eigenvalue. We start with u1T = [1 0 0], so that x1 = Mu  1 2 = 0 0 The computations involved in the inverse iteration are tabulated below. k

xk   2 0 0

1

 1.17 −0.56 −0.68   1.190 −0.521 −0.667 

2

3

u¯ k+1  14.20 −3.41 −8.23   12.10 −2.64 −6.78   12.09 −2.65 −6.79 

x¯ k+1 



28.40 −13.65 −16.45   24.19 −10.57 −13.56   24.18 −10.60 −13.58

T u¯ k+1 xk

T u¯ k+1 x¯ k+1

ρk+1

28.4

585.2

0.0485

20.25

412.5

0.0491

20.29

412.3

0.0492

xk+1  1.17 −0.56 −0.68   1.190 −0.521 −0.667 

576

Dynamics of structures Convergence has been achieved at this stage. We therefore have δ = 0.0492 λ=µ+δ = 1.67 + 0.0491 = 1.719 u¯ 4 φ2 = T 1/2 [u¯ 4 x¯ 4 ]   0.5952 = −0.1305 −0.3342

In carrying out the determinant search for the third eigenvalue, we use the deflated polynomial p(3) =

p (µ − 0.5)(µ − 1.719)

For our two initial estimates of λ3 we choose 2.0 and 2.5, respectively. The remaining computations are tabulated as follows.

k

µk−2

p(3) k−2

µk−1

p(3) k−1

µk

p(3) k

1

2.0

14.24

2.5

10.24

3.781

0

We have converged exactly on the third eigenvalue. This is to be expected because p(3) is a linear function of µ, and secant iteration should therefore lead to the true zero in one step. To obtain the eigenvector, we may carry out an inverse iteration with a shift of 3.75. The final result are δ = 0.031 λ=µ+δ = 3.75 + 0.031 = 3.781   0.1441 φ3 = −0.3286 0.5131

11.7

NUMERICAL SOLUTION OF COMPLEX EIGENVALUE PROBLEM

11.7.1 Eigenvalue problem and the orthogonality relationship A majority of eigenvalue problems in engineering analysis involve matrices that are symmetric and positive definite. Occasionally, cases arise where eigenvalues of a real

Numerical solution of the eigenproblem

577

unsymmetric or even a complex matrix may be required. Systems with nonproportional damping discussed in Section 10.8 belong to the former category. The eigenvalue problem is, in such a case, given by Bv = sAv

(11.151)

where A and B are the following matrices, each of order 2N:     0 −M −M 0 A= B= −M −C 0 K and vector v is given by   u˙ v= u

(11.152a)

(11.152b)

It should be noted that although both A and B are symmetric and real, they are not positive definite. The eigenvalue problem of Equation 11.151 can be expressed in the alternative form Ev = sv

(11.153)

where E = A−1 B is given by   −M−1 C −M−1 K E= I 0

(11.154)

Matrix E is real but unsymmetric. For an underdamped system the 2N eigenvalues, s, and eigenvectors, v, are both complex. Furthermore, they both appear in complex conjugate pairs, and for a system that exhibits exponentially decaying free vibrations, the real part of each eigenvalue is negative. Equation 11.151 can also be expressed in the form B−1 Av =

1 v s

(11.155)

or Dv = γ v

(11.156)

where D = B−1 A and γ = 1/s. Matrix D, called the dynamic matrix, is given by   0 I (11.157) D= −K−1 M −K−1 C in which submatrix K−1 M would be recognized as the dynamic matrix for the undamped case. We denote the jth eigenvalue sj as sj = αj + iβj

(11.158)

578

Dynamics of structures

and its complex conjugate s¯j as s¯j = αj − iβj

(11.159)

where αj and βj are both real numbers. The corresponding eigenvectors are denoted by vj = ξ j + iηj

(11.160a)

v¯ j = ξ j − iηj

(11.160b)

As proved in Section 10.8, the eigenvectors are orthogonal with respect to both A and B, so that vjT Avk = 0

j = k

(11.161a)

vjT Bvk = 0

j = k

(11.161b)

The orthogonality relationships hold even between two eigenvectors that are complex conjugated of each other, since their eigenvalues are different. It is useful to express the orthogonality relationships in terms of the real and imaginary parts. Thus Equation 11.161a is written as (ξ Tj + iηTj )A(ξ k + iηk ) = 0

(11.162)

and since eigenvector vk is orthogonal also to the conjugate of eigenvector vj , (ξ Tj − iηTj )A(ξ k + iηk ) = 0

(11.163)

Expansion of Equations 11.162 and 11.163 gives ξ TjAξ k − ηTjAηk = 0 ηTjAξ k + ξ TjAηk = 0 ξ TjAξ k + ηTjAηk = 0

(11.164)

ηTjAξ k − ξ TjAηk = 0 Combining the first and third, and second and fourth of Equations 11.164, we get the family of orthogonality relationships ξ Tj Aξ k = 0 ηTj Aηk = 0 ξ Tj Aηk = 0

j = k

ηTj Aξ k = 0 A similar set of results can be expressed with respect to matrix B.

(11.165)

Numerical solution of the eigenproblem

579

Now since an eigenvector is also orthogonal to its own conjugate, we have (ξ Tj + iηTj )A(ξ j − iηj ) = 0

(11.166)

and therefore ξ Tj Aξ j = −ηTjAηj ηTj Aξ j = ξ TjAηj

(11.167)

11.7.2 Matrix iteration for determining the complex eigenvalues The orthogonality relationships proved above are useful in developing a numerical technique for the solution of the aforesaid complex eigenvalue problem by a process of inverse iteration similar to that used in Section 11.5.2. It is, in fact, possible to start with an initial trial vector that is real and work entirely with real numbers to compute both the real and the imaginary parts of eigenvalues and eigenvectors. To develop (0) (0) the procedure, let us begin the iteration with a complex trial vector p1 + iq1 . The superscript (0) indicates an initial trial vector and the subscript 1 refers to the fact that the iteration are for the first eigenvector. For simplicity, we omit the subscript in the following discussion. Also, we will express the inverse eigenvalue γ in terms of its real and imaginary components, γ = µ + iν

(11.168)

The iteration involves repeated multiplication of the trial vector by matrix D. A single step in this iteration process is given by D(p(n) + iq(n) ) = (µ + iν)(p(n) + iq(n) ) = p(n+1) + iq(n+1)

(11.169)

Expansion of Equation 11.169 gives Dp(n) = µp(n) − νq(n) = p(n+1)

(11.170a)

Dq(n) = µq(n) + νp(n) = q(n+1)

(11.170b)

By eliminating q(n) from Equations 11.170a and 11.170b, we get an expression for q(n+1) in terms of p(n) and p(n+1) νq(n+1) = (µ2 + ν 2 )p(n) − µp(n+1)

(11.171)

Next we write Equation 11.170a at step n + 1, so that Dp(n+1) = µp(n+1) − νq(n+1) = p(n+2)

(11.172)

580

Dynamics of structures

or νq(n+1) = µp(n+1) − p(n+2)

(11.173)

Substitution of Equation 11.173 in Equation 11.171 gives p(n+2) − 2µp(n+1) + (µ2 + ν 2 )p(n) = 0

(11.174)

Equation 11.174 can be viewed as a recursion equation that relates vector p at three successive iteration steps. Thus the next equation in the sequence will be p(n+3) − 2µp(n+2) + (µ2 + ν 2 )p(n+1) = 0

(11.175)

Equations 11.174 and 11.175 hold for every element of vector p. By solving these equations for any one of the elements of p, we obtain the real part of the eigenvalue: µ=

1 p(n) p(n+3) − p(n+1) p(n+2) 2 p(n) p(n+2) − p(n+1) p(n+1)

(11.176)

in which p denotes one of the elements of vector p. The same two equations, Equations 11.174 and 11.175, also give the squared amplitude (µ2 + ν 2 ) of the complex eigenvalue µ2 + ν 2 =

p(n+1) p(n+3) − p(n+2) p(n+2) p(n) p(n+2) − p(n+1) p(n+1)

(11.177)

With a sufficient number of iterations, µ will converge to the real part of the inverse eigenvalue, γ , having the highest amplitude and µ2 + ν 2 will converge to the squared amplitude of that eigenvalue. Simultaneously, p(n) will converge to the real part of the eigenvector. The imaginary part of the eigenvector can be obtained from Equation 11.170a: q(n) =

µ (n) 1 (n+1) p − p ν ν

(11.178)

To obtain the eigenvalue s = α + iβ, we use the relationships µ + ν2 ν β=− 2 µ + ν2 α=

µ2

(11.179a) (11.179b)

Convergence cannot, in this case, be measured by comparing the elements of two successive values of the trial vector. Instead, comparison must be made between two successive values of µ or of the squared amplitude µ2 + ν 2 . The procedure just described gives the eigenvalue with the lowest amplitude, or the inverse eigenvalue with highest amplitude, as well as the corresponding eigenvector. To obtain a higher eigenvalue, the eigenvalues already determined must be swept off from the trial vector. This is achieved by writing the orthogonality relationships between a

Numerical solution of the eigenproblem

581

trial vector for determining the nth eigenvalue and the n − 1 eigenvectors determined previously. Let us assume that the eigenvector (ξ 1 + iη1 ) has already been obtained and the second eigenvalue and eigenvector are to be determined. Denoting the trial vector for the second eigenvector by p2 + iq2 , we have ξ T1Ap2 = 0 ηT1Aq2 = 0

(11.180)

ξ T1Aq2 = 0 ηT1Ap2 = 0

Since we carry out our iterations on a real vector, we need only the first and fourth of Equations 11.180. Observing that there are two constraint equations, we can assign arbitrary value to all elements of the trial vector except two, say the first two. The first two elements are then selected so that the complete trial vector satisfies the two constraint equations. The procedure is entirely similar to that described in Section 11.5.2, in which the constraint conditions were used to obtain a sweeping matrix. Premultiplication of the trial vector by this matrix resulted in the first eigenvector being swept off. In the present case, the sweeping matrix obtained from the two constraint equations will have two columns of zeros. This is to be expected because in calculating the first eigenvector, we have, in fact, obtained a complex conjugate pair rather than a single vector. The use of the fourth of Equation 11.180 requires the computation of η1 , the imaginary part of the first eigenvector. Vector η1 = q(n) was probably calculated from Equation 11.178 during determination of the first eigenvector. However, as an alternative, we can use the following procedure, which avoids the use of η1 . (n) Noting that ξ 1 = p1 , we express the first of Equations 11.180 as (n)

{p1 }TAp2 = 0

(11.181) (n)

In a similar manner, noting that η1 = q1 , we write the fourth of Equation 11.180 as (n)

{q1 }TAp2 = 0

(11.182) (n)

Substitution for q1 from Equation 11.178 into Equation 11.182 gives µ (n) T 1 (n+1) {p } Ap2 − {p1 }TAp2 = 0 ν 1 ν

(11.183)

The first term in Equation 11.183 is zero because of the relationship in Equation 11.181. Hence (n+1) T

{p1

} Ap2 = 0

(11.184)

Equations 11.181 and 11.184 represent the two constraints that p2 must satisfy.

582

Dynamics of structures

Example 11.9 For the two-degree-of-freedom system described in Example 10.3, obtain the complex mode shapes and frequencies by the iteration method.

Solution The mass, stiffness, and damping matrices are given in Example 10.3. The eigenvalue problem is defined by Equation 11.151 Bv = sAv where 

0  0  A= −2.0 0

0 0 0 −1.0

−2.0 0 −0.4 0.05

0 −1.0 0 0

0 0 3.0 −1.0

 0 −1.0   0.05 −0.2

(a)

and 

−2.0  0  B= 0 0

 0 0   −1.0 1.0

(b)

For solution by the inverse iteration method, we express the problem in the form Dv = γ v

(c)

where the dynamic matrix D = B−1 A is given by 

0  0 D= −1.0 −1.0

0 0 −0.5 −1.5

1.0 0 −0.175 −0.125

 0 1.0   −0.075 −0.274

(d)

Iteration is begun with a trial vector given by pT = [1 obtained in the iteration process are tabulated as follows. n=0

n=1

n=2

n=3

n=4

n=5

1.0000

1.0000 −1.7500 −0.9763 3.4821

0.6115

1.0000

1.0000 −2.9000 −1.4837 6.6301

1

1

1]. Successive vectors

n=6

n=7

n=8

−6.9749

1.1697

13.4550

0.9433 −13.7630

2.6302

26.7500 −14.1530

1.1697 13.4550

−6.8456 −24.5700

1.0000 −1.7500 −0.9763

3.4821 0.6115

−6.9749

1.0000 −2.9000 −1.4837

6.6301 0.9433 −13.7630

n=9 −6.8456

2.6302 26.7500 −14.1530 −48.8320

After the first three new vectors have been calculated, Equations 11.176 and 11.177 are used to obtain estimates for µ and µ2 + ν 2 , respectively. The two equations can be evaluated using any one of the four elements of the iteration vector. In the following we tabulate the values of µ and µ2 + ν 2 obtained by using the third elements of the iteration vectors.

Numerical solution of the eigenproblem

n

p(n) 3

µ (Eq. 11.176)

µ2 + ν 2 (Eq. 11.177)

0

1.0000

−0.21958

1.7448

1

−1.7500

−0.16527

1.8054

2

−0.9763

−0.18393

1.9385

3

3.4821

−0.16905

1.9437

4

0.6115

−0.17092

1.9863

5

−6.9749

−0.16810

1.9854

6

1.1697

−0.16761

1.9967

7

13.4550

8

−6.8456

9

−24.5700

583

For the purpose of this example we accept the convergence achieved at the sixth iteration (n = 6). We can now use Equations 11.179a and 11.179b to obtain the real and imaginary parts of the eigenvalue s. Thus µ = −0.16761 ν = ±1.4031 µ = −0.08394 µ2 + ν 2 −ν = ∓0.70269 β= 2 µ + ν2

(e)

α=

(f)

The real part of the eigenvector is taken from the first table above, corresponding to n = 6:  −6.9749 −13.7630  ξ=  1.1697 2.6302 

(g)

The imaginary part of the eigenvector can be calculated from Equation 11.178: µ (6) 1 (7) p − p ν ν

(h)

 −0.00046122  −0.23053  η = ±  −9.7290 −19.380

(i)

η = q(6) = which gives 

Vectors ξ and η can be normalized with respect to matrix A so that (ξ T + iηT )!A(ξ + η) = 1

(j)

584

Dynamics of structures

The normalized vectors are 



0.19471

 0.37928   ξ =  −0.24165

(k)

−0.48973 

 0.14985 0.30209   η = ±  0.24648

(l)

0.48453 To obtain the second eigenvectors, we note that its real part must satisfy the two constraints, Equations 11.181 and 11.184. Substituting for p(6) and p(7) and A in these equations, we get 2.3393 p1,2 + 2.6302 p2,2 − 13.613 p3,2 − 13.296 p4,2 = 0

(m)

26.909 p1,2 + 26.750 p2,2 + 6.3836 p3,2 + 7.3075 p4,2 = 0

In the constraint equations above, the first subscript on p denotes the element number in vector p, and the second subscript signifies that the iteration vector is for the second mode. With two constraint equations, we are free to assign arbitrary values to any two of the four elements of p2 ; the other two are then selected so that the constraint equations are satisfied. Let us choose to express elements p1 and p3 in terms of p2 and p4 . Then denoting the purified vector by p˜ 2 , we can express the constrain relationship as p˜ 2 = Sp2

(n)

where S is the sweeping matrix given by 

0

 0 S= 0 0

−0.99919

0

1 −0.021504

0 0

0

−0.38311



 0   −0.98323 1

0

(o)

The dynamic matrix for calculating the second set of eigenvalues and eigenvectors is obtained from D2 = DS  0 0  = 0 0

0.021504

0

0

0

0.49543

0

 −0.98323  1   0.13538

−0.50359

0

−0.11379

Iterations are again begun with a trial vector given by pT2 = [1 obtained in the iteration process are tabulated as follows.

(p)

1

1

1]. Successive vectors

Numerical solution of the eigenproblem

n=0

n=1

1.0000

n=2

n=3

−0.96172

0.62843

0.41272

585

n=4 −0.36337

1.0000

1.00000

−0.61728

−0.43326

0.36010

1.0000

0.63081

0.41137

−0.36447

−0.16590

1.0000

−0.61728

−0.43326

0.36010

0.17717

Estimates of µ and µ2 + ν 2 are now obtained from Equations 11.176 and 11.177, respectively, and are tabulated as follows. Again, we use the third element of the iteration vector in our calculations. p(n) 3,2

n

µ2 (Eq. 11.176)

µ2 + ν 2 (Eq. 11.177)

0

1.00000

−0.056893

0.503495

1

−0.61728

−0.056893

0.503495

2

−0.43326

3

0.36010

4

0.17717

Convergence has been achieved at n = 1. The corresponding values of µ, ν, α, and β are obtained as follows: µ2 = −0.056893 ν2 = ±0.70729 α2 = −0.11300

(q)

β2 = ∓1.4048 The real part of the eigenvectors is equal to the iteration vector for n = 1:  −0.96172  1.00000  ξ2 =   0.63081 −0.61728 

(r)

The imaginary part is obtained from Equation 11.178  −0.81115  0.79230  η2 = ±  −0.63306 0.66221 

(s)

The normalized eigenvectors are given by  0.36533 −0.38012  ξ2 =  −0.24526 0.24021 

(t)

586

Dynamics of structures  0.31610 −0.30905  η2 = ±   0.24001 −0.25126 

(u)

The results obtained in this example compare well with the more exact values given in Example 10.3. The small differences exist because, in the present case, iterations for the first set of vectors were terminated early at n = 6.

11.8

SEMIDEFINITE OR UNRESTRAINED SYSTEMS

11.8.1 Characteristics of an unrestrained system It can be shown that for a stable system that is adequately restrained against rigid-body motion, the stiffness matrix is positive definite. This follows from the fact that when such a system is subjected to arbitrary displacement, the resulting strain energy stored in the system is always positive. If the arbitrary displacement vector is u, the forces required to cause such displacements are f = Ku. The work done by these forces as they gradually increase from zero to their full value is stored as strain energy V, which is given by V= =

1 T u f 2 1 T u Ku 2

(11.185)

Since V is positive for any arbitrary u, the term uT Ku is also positive. By definition, therefore, K is a positive definite matrix. In systems that are not properly restrained, rigid-body displacements can take place without the application of any force. Thus, denoting a possible rigid-body displacement by ur , we have fr = Kur = 0

(11.186)

For nonzero ur , Equation 11.186 can be satisfied provided only that K is singular. It follows from Equation 11.186 that the strain energy stored in a rigid-body displacement is also zero, 1 T u fr 2 r 1 = urT Kur 2 =0

Vr =

(11.187)

A matrix K which has the property that for arbitrary u, the quadratic form uT Ku is either positive or zero is called a semidefinite matrix. Therefore, an unrestrained

Numerical solution of the eigenproblem

587

system that permits rigid-body displacement has a semidefinite stiffness matrix. Such a matrix is also singular. To illustrate these concepts, consider the system of Example 11.6. The stiffness matrix of the system is given by   k1 −k1 0 K = −k1 k1 + k2 −k2  (11.188) 0 −k2 k2 Suppose that we wish to impose arbitrary displacements u1 , u2 , and u3 along coordinates 1, 2, and 3, respectively. External forces must be applied to the system to maintain the specified displacements. These forces will perform work as they undergo their corresponding displacements, and the work done will be stored in the system as stain energy. Now consider a special displacement pattern given by u1 = u2 = u3 . The forces required to cause such a displacement can be found by substitution in Equation 11.186 and are easily seen to be zero. The corresponding strain energy is also therefore zero. The special displacement pattern u1 = u2 = u3 will be recognized as a rigid-body motion of the system. In fact, since the system is constrained to move in the plane of the paper, this is the only possible rigid-body motion. As would be expected, the stiffness matrix given by Equation 11.188 is singular, the second row of the matrix being a linear combination of the first and the last rows. The degree of singularity of the matrix is 1, corresponding to the one possible rigid-body displacement pattern. It is of interest to note that a rigid-body displacement shape is also an eigenvector of the unrestrained system. For this statement to be true, the displacement vector should satisfy Equation 10.4 or equivalently, Equation 10.7. Thus Kur = ω2 Mur

(11.189)

Since Kur is zero, Equation 11.189 can be satisfied by choosing ω = 0. This proves the statement that ur is an eigenvector or a mode shape, the corresponding eigenvalue or frequency being zero. The number of zero eigenvalues of Equation 11.189 is equal to the number of rigid-body modes, that is, equal to the degree of singularity of matrix K. Rigid-body displacement shapes are also referred to as rigid-body modes. They have the same properties as other mode shapes. In particular, they satisfy the orthogonality relationships given by Equations 10.14 and 10.15. A system can, of course, have more than one rigid-body mode. In the most general case, up to six rigid-body modes are possible. For example, a spacecraft or an aeroplane in flight has all six possible rigid-body modes, three translations and three rotations, one along each of the three axes.

11.8.2 Eigenvalue solution of a semidefinite system As pointed out in the foregoing paragraphs, an unrestrained or semidefinite system has a singular stiffness matrix. This implies that in evaluating the eigenvalue of such a system, all those methods that rely on obtaining an inverse of the stiffness matrix cannot

588

Dynamics of structures

be used. Thus the linearized eigenvalue problem cannot be reduced to a symmetric form by using Equation 11.15b. Also, it is not possible to use inverse iteration, the subspace iteration, or the Lanczos iteration methods to obtain an eigenvalue solution. As noted in earlier sections of this chapter, in many cases, an iteration method may be the most efficient method of finding the mode shapes and frequencies of a system. Such is the case, for example, when the system is large and only the first few mode shapes and frequencies and required. Iteration methods cannot, however, be applied directly to the eigensolution of unrestrained system and special procedures must therefore be used in such cases. Three alternative procedures exist and are described in the following paragraphs. Addition of a small fictitious stiffness The singular stiffness matrix of the unrestrained system can be made nonsingular by adding a small but fictitious stiffness along an adequate number of degrees of freedom. This is equivalent to adding restraining springs along one or more degrees of freedom so that the rigid-body motions are inhibited. To account for the presence of such springs, appropriate stiffness terms must be added to the diagonal of the original stiffness matrix. Addition of restraining springs while making the stiffness matrix nonsingular also modifies the system. However, if the restraining springs have a very low stiffness, the frequencies and mode shapes of the modified system will be very nearly equal to those of the original system. Example 11.10 illustrates the application of this method to the eigensolution of system shown in Figure E11.6. Example 11.10 Obtain the eigenvalues and eigenvectors of the system shown in Figure E11.6 by adding a fictitious stiffness to degree-of-freedom 1.

Solution The semidefinite system of Figure E11.6 is modified by adding a spring of stiffness k0 along degree-of-freedom 1. The modified system is shown in Figure E11.10. If we select k0 = 0.01, the stiffness matrix of the system becomes 

2.01 K = −2 0

−2 6 −4

 0 −4 4

The mass matrix remains unchanged.

Figure E11.10 Semidefinite system restrained by a fictitious spring.

Numerical solution of the eigenproblem

589

An inverse iteration method as described by Equations 11.73 through 11.76 is used to obtain the eigenvalues and eigenvectors. The steps involved are presented below in brief. For each mode the first trial vector u is selected so that uT = [1 1 1]. First mode. Iteration begins with x1 = Mu. The computations are tabulated as follows.

k

u¯ k+1

xk   2   4 2

1





 0.70501   1.41530 0.70810

2

x¯ k+1

 800.00   803.0  803.50



  282.84   283.90 284.08



xk+1 

1600.0   3212.0 1607.0 

565.68

  1135.60 568.16



ρk+1 

0.70501   1.4153  0.70810 

0.70501

0.0012463



  1.4153  0.70810

0.0012463

Convergence has been achieved at this stage. The eigenvalue is 0.0012463, as compared to the exact value of 0.0. The eigenvector is given by u¯ 3 ; when normalized so that the first element is equal to 1, it becomes 

 1   q1 = 1.0037 1.0044

compared to the rigid-body mode given by q1T = [1 1 1]. As an alternative, the eigenvector can be mass orthonormalized using Equation 11.76. This gives 



0.35251

  φ1 = 0.35383 0.35405

Second mode.

To obtain the second eigenpair a sweeping matrix is constructed using

S1 = I − φ1 φ1T M  0.75148  = −0.24945

 0.49891 −0.24961  0.49922 −0.25055 −0.24961 −0.50109 0.74930

Iteration for the second mode begin with x1 = MS1 u. Also, Equation 11.97 is used in place of Equation 11.74. The computations are tabulated as follows.

590

Dynamics of structures

k 1

2

3

4

u¯ k+1

xk 



0.0059318 −0.0031066 −0.0028013   1.20050 −0.56285 −0.63277   1.1950 −0.53537 −0.65474   1.19280 −0.52507 −0.66286



x¯ k+1 

0.0023930 −0.00056097 −0.0025226   0.48840 −0.10941 −0.26760   0.48771 −0.10735 −0.27103   0.48740 −0.10656 −27228



xk+1 

0.0047060 −0.0022439 −0.0025226   0.97681 −0.43762 −0.53520   0.97542 −0.42938 −0.54206   0.97480 −0.42626 −0.54456



ρk+1 

1.2005 −0.56285 −0.63277   1.1950 −0.53537 −0.65474   1.19280 −0.52507 −0.66286   1.19200 −0.52122 −0.66587

1.2251

1.2231

1.2228

1.2228

Convergence has been achieved at this stage. The second eigenvalue is 1.2228; the corresponding eigenvector is given by u¯ 5 . When normalized so that the first element is equal to 1, the eigenvector becomes  1.00000 q2 = −0.21863 −0.55863 

In comparison, the more precise value obtained in Example 11.6 was  1.0000 q2 = −0.2192 −0.5616 

When mass orthonormalized, the second eigenvector becomes  0.59577 φ2 = −0.12981 −0.33371 

It is of interest to note that the selection of uT = [1 1 1] as the initial trial vector was not a very good choice in this case, because the trial is very close to the first mode shape. As a result, when the first mode shape is swept off from it, the elements of the resulting vectors are very small, as shown by the small values in x1 = MS1 u. This does not, however, pose any special problem with the precision available on a computer. Third mode. The sweeping matrix needed to obtain the third mode is given by S2 = I − φ1 φ1T M − φ2 φ2T M = S1 − φ2 φ2T M   0.041607 −0.18956 0.14801 = −0.094778 0.43182 −0.33718 0.14801 −0.67437 0.52658

Numerical solution of the eigenproblem

591

Iterations are begun with x1 = MS2 u. The computations are tabulated as follows. k

1

2

3

u¯ k+1

xk  −3



0.13251 × 10 −0.57933 × 10−3  0.44706 × 10−3   0.28848 −1.31430 1.02620   2.8846 −1.13143 1.0262



x¯ k+1  −4

0.24091 × 10 −0.42044 × 10−4  0.69722 × 10−4   0.043901 −0.10011 0.15645   0.043885 −0.10013 −0.15643



xk+1  −4

0.38554 × 10 −0.17579 × 10−3  −0.13727 × 10−3   0.037919 −0.40057 0.31278   0.087919 −0.40057 0.31278



ρk+1 

0.28848 −1.31143 1.02620   0.28846 −1.13143 1.0262   0.28846 −1.13143 1.0262

3.2840

3.2810

3.2810

Iterations have converged; the eigenvalue is 3.2810 as compared to the more precise value of 3.271. The eigenvector is given by u¯ 4 . When normalized so that the first element is 1, it becomes  1.0000 q3 = −2.2816 3.5645 

as compared to the more precise value in Example 11.6  1.000 q3 = −2.280 3.561 

On mass orthonormalizing the eigenvector, we obtain  0.14399 φ3 = −0.32851 0.31325 

It is clear from the solution in Example 11.10 that the fictitious spring could have been added to any one or more of the three degrees of freedom. It is also apparent that a certain approximation is involved in the solution because of the addition of one or more fictitious springs. However, with the numeric precision that can be obtained in modern computers, the stiffness of the fictitious springs can be kept very small, so that the eigenvalues and eigenvectors of the modified system will, for all practical purposes, be equal to those of the original system. Also, the computer algorithm can be designed so that whenever during reduction of the stiffness matrix a zero is encountered on the diagonal, it is replaced by a small positive value and the reduction resumed from that point. This automates the procedure of adding springs to restrain rigid-body motions.

592

Dynamics of structures

Vector iteration with shift As explained in Section 11.5.3, introduction of a shift µ in the origin of the eigenvalue axis modifies the eigenvalue problem to that given by Equation 11.99a (K − µM)q = δMq

(11.99a)

where λ, the eigenvalue of the original problem, is related to δ by Equation 11.98, λ = µ + δ. The modified stiffness matrix (K − µM) is, in general, nonsingular (if it is not, another value can be selected for µ), so that the vector iteration method can be applied to the solution of a modified problem. The eigenvectors remain unchanged under the shift; the eigenvalue λ can be obtained from δ by using Equation 11.98. Iteration with shifts does not involve any approximation; both the eigenvalues and eigenvectors are obtained accurately. It may also be noted that when the mass matrix is diagonal, a shift is equivalent to adding springs along all degrees of freedom, the stiffness of a spring being equal to −µ times the corresponding term on the diagonal of the mass matrix. The application of iteration with shift to the eigensolution of a semidefinite system is illustrated by Example 11.6. Sweeping of rigid-body modes The rigid-body modes of an unrestrained or partially restrained system can usually be obtained by inspection. The condition that the remaining modes of the system be orthogonal to the rigid-body modes can therefore be used to set up as many constraint equations as there are rigid-body modes. Thus if of a total of N mode shapes, s is the number of rigid-body modes and r the number of remaining modes, there exist s constraint equations. For an eigensolution of the unrestrained system these equations are used to obtain a coordinate transformation in which the s + r coordinates are expressed in terms of the r independent coordinates. A reduced eigenvalue problem is then formed in the transformed coordinates. Since the rigid-body modes have been removed in the process of transformation, the reduced system is positive definite and its eigenvalues and eigenvectors can be obtained by any iteration process. The procedure is entirely equivalent to that described in Section 11.5.2 for sweeping off the mode shapes already determined. The r remaining modes satisfy the constraint equations expressed by Equation 11.90, 11.91, and 11.92. If one of the remaining mode shape u is partitioned as shown in Equation 11.93, the constrain condition can be expressed as u=

 T  (s Ms )−1 (Ts Mr ) ur I

= Tur

(11.190a) (11.190b)

Equation 11.190 can be viewed as a coordinate transformation in which transformation matrix T is of size N by r and vector ur is of size r. The transformation is used to form an eigenvalue problem in the reduced space of size r. This gives Kur = λMur

(11.191)

Numerical solution of the eigenproblem

593

where K = TT KT M = TT MT

(11.192)

and both K and M are of size r by r. Solution of the eigenvalue Equation 11.191 will provide the remaining r eigenvalues and the corresponding eigenvectors. It should be noted that an eigenvector of Equation 11.191 will have only r elements. The corresponding full eigenvector of the original problem of size N can, however, be obtained by using Equation 11.190. For a system of significant size, the number of rigid-body modes is small compared to the total number of degrees of freedom. As a result, the size of reduced stiffness matrix K or of the reduced mass matrix M is not much smaller than that of the original matrices. On the other hand, the transformation given by Equation 11.192 destroys the bandedness of the stiffness and mass matrices. Because of this, the method of sweeping off the rigid-body modes is computationally inefficient in comparison to the other two methods described in this section. Example 11.11 Obtain the eigenvalues and eigenvectors of the unrestrained system of Example 11.6 by sweeping off the rigid-body modes.

Solution It is easily seen that the system has one rigid-body mode, corresponding to a horizontal translation in the plane of the paper. This mode is given by   1 q1 = 1 1

(a)

The corresponding eigenvalue is λ1 = 0. Let one of the two remaining modes be represented by u, where uT = [u1 u2 u3 ]. Since this mode should be orthogonal to the rigid-body mode already determined, we have q1T Mu = 0 or 2u1 + 4u2 + 2u3 = 0

(b)

Equation b can be expressed in the form of a coordinate transformation as follows:    −2 u1 u2  =  1 0 u3

 −1   u 0 2 u3 1

(c)

or as u = Tur

(d)

594

Dynamics of structures

where 

−2 T= 1 0

 −1 0 1

(e)

We now form a reduced eigenvalue problem as follows: K = TT KT   22 2 = 2 6

M = TT MT   12 4 = 4 4 Kur = λMur

(f)

(g) (h)

We note that although K was singular, K is not. The eigenproblem of Equation h can therefore be solved by inverse iteration. The lower of the two eigenvalues obtained from Equation h is 1.2192. This is the second eigenvalue of the original system. Thus λ2 = 1.2192

(i)

The corresponding eigenvector in the reduced space is  u2 =

1.0000 2.5617

 (j)

We now use the coordinate transformation given by Equation d to obtain the full eigenvector.   −2 −1  1.0000 0 q˜ 2 =  1 2.5617 0 1   −4.5617 =  1.0000 2.5617 

On dividing through by the first element, we obtain the normalized vector  1.0000 q2 = −0.2192 −0.5615 

which is identical to that obtained in Example 11.6.

(k)

Numerical solution of the eigenproblem

595

On continuing the inverse iteration procedure, we get λ3 = 3.2807  1.0000 u3 = −1.5615 

(l)



−0.4387

  q˜ 3 =  1.0000 −1.5615

(m)

and 



1.0000

  q3 = −2.2805

(n)

−3.5610

11.9

SELECTION OF A METHOD FOR THE DETERMINATION OF EIGENVALUES

The discussion in previous sections of this chapter shows that several alternative methods are available for the solution of an eigenproblem. Which method is most effective in a particular case depends on the type of solution desired and the characteristics of the matrices involved. The advantages and limitations of the methods presented were pointed out as each method was discussed. It is instructive to summarize these and to show how they affect the selection of a particular technique for the solution of the problem in hand. Often, the procedure that is most effective for the problem to be solved will involve a combination of two or more different methods. The transformation methods are most useful when the matrices are of comparatively small order and are more or less fully populated or have a large bandwidth. They are not particularly appropriate when the matrices involved are of a large order and only a few eigenvalues need to be determined. In all cases, the eigenvector corresponding to a computed eigenvalue is best determined by a process of inverse iteration with shift equal to the eigenvalue. The Jacobi transformation method is simple and stable. It is appropriate for calculating the eigenvalues of a standard symmetric matrix. The matrix need not be positive definite and the eigenvalues being sought may be negative, zero, or positive. All eigenvalues are obtained simultaneously. The generalized Jacobi method is suitable for the solution of a linearized eigenvalue problem of small order, particularly when the matrices involved are banded. The Householder transformation matrix operates on a standard symmetric matrix. It must be combined with the QR method for the determination of eigenvalues. The eigenvectors are then obtained by using inverse iteration with shifts. For standard symmetric matrices, the combination of Householder and QR method is, in general, more efficient than the Jacobi method. As in the standard Jacobi method, the matrix need

596

Dynamics of structures

not be positive definite, and negative, zero, or positive eigenvalues may be obtained. Also, all eigenvalues are obtained simultaneously. Iteration methods are most effective when the matrices involved are of a large order and only a few eigenvalues are desired. They may be applied directly to linearized eigenvalue problems without the need of transforming the latter to a standard form. Direct iteration gives the eigenvalues with the largest magnitude. For the solution of a linearized eigenproblem by direct iteration, the mass matrix must be positive definite. Inverse iteration leads to the lowest eigenvalue. For solution of a linearized eigenproblem by inverse iteration, the stiffness matrix must be positive definite. When using an iteration method to determine an eigenvalue other than the most dominant or the least dominant, the eigenvectors already determined must be swept off from the trial vector by using Gram–Schmidt orthogonalization. The accuracy of subsequent eigenvectors depends on the precision with which the previous eigenvectors have been determined. Therefore, unless high numerical precision is maintained, the methods may not be suitable for determining eigenpairs other than the few most dominant or least dominant. When more than a few lower-order eigenpairs are required, inverse iteration with shifts is suitable. The shifting procedure is, however, not very effective with forward or direct iteration. For problems of large size where only a few eigenpairs are required, Lanczos iteration, the determinant search method, and subspace iteration are most effective. The determinant search method is used in conjunction with inverse iteration with shifts and Sturm sequence check. Subspace iteration is used together with Jacobi transformation and Sturm sequence check. SELECTED READINGS Bathe, K. J., Finite Element Procedures, Prentice-Hall, Englewood Cliffs, NJ, 1996. Bathe, K.J., and Wilson, E.L., “Solution Methods for Eigenvalue Problems in Structural Mechanics’’, International Journal for Numerical Methods in Engineering, Vol. 6, 1973, pp. 213–226. Bathe, K.J., and Wilson, E.L., “Eigensolutions of Large Structural Systems with Small Bandwidth’’, Journal of Engineering Mechanics Division, ASCE, Vol. 99, 1973, pp. 467–479. Bathe, K.J., and Wilson, E.L., Numerical Methods in Finite Element Analysis, Prentice Hall, Englewood Cliffs, 1976. Francis, J.G.F., “The QR Transformation Parts I and II’’, Computer Journal, Vol. 4, 1961–62, pp. 265–371 and 332–345. Goldstine, H.H., Murray, F.J., and Von Neumann, J., “The Jacobi Method for Real Symmetric Matrices’’, Journal of the Association for Computing Machinery, Vol. 6, 1959, pp. 59–96. Gupta, K.K., “Solutions of Eigenvalue Problems by the Sturm Sequence Method’’, International Journal for Numerical Methods in Engineering, Vol. 4, 1972, pp. 379–404. Householder, A.S., Principles of Numerical Analysis, McGraw-Hill, New York, 1963. Jennings, A., Matrix Computation for Engineers and Scientists, 2nd Edition, John Wiley, Chichester, U.K., 1992. Jennings, A., “Eigenvalue Methods and the Analysis of Structural Vibration’’, in Sparse Matrices and Their Uses, (Ed. I.S. Duff), Academic Press, London, 1981, pp. 109–138. Ortega, J.M., and Kaiser H.F., “The LLT and QR Methods for Symmetric Tridiagonal Matrices’’, Computer Journal, Vol. 6, 1963–64, pp. 99–101. Parlett, B.N., The Symmetric Eigenvalue Problem, Prentice-Hall, Englewood Cliffs, 1980.

Numerical solution of the eigenproblem

597

Parlett, B.N., and Scott S.N., “The Lanczos Algorithm with Selective Orthogonalization’’, Mathematics of Computation, Vol. 33, 1979, pp. 217–218. Scott, D.S., “The Lanczos Algorithm,’’ in Sparse Matrices and Their Uses, (Ed. I.S. Duff), Academic Press, London, 1981, pp. 139–159. Weaver, W. Jr., and Yoshida D.M., “The Eigenvalue Problem for Bounded Matrices,’’ Computers and Structures, Vol. 1, 1971, pp. 651–664. Wilkinson, J.H., The Algebraic Eigenvalue Problem, Clarendon Press, Oxford, U.K., 1965.

PROBLEMS 11.1

Find the mass and the flexibility matrices of the simply supported uniform beam shown in Figure P11.1 corresponding to the 3 coordinate direction indicated. The eigenvalue problem for the flexural vibrations of the beam can be expressed as K−1 Mq = Dq = γ q

1 q λ

Figure P11.1 Show that for the given system D is a symmetric matrix. Using standard Jacobi method, obtain the 3 natural frequencies of the system. 11.2

By inverting the flexibility matrix determined in Problem 11.1, obtain the stiffness matrix of the beam. Compute the three frequencies of the beam by using the generalized Jacobi method.

11.3

Determine the frequencies and mode shapes of the beam in Figure P11.1 by the inverse iteration method.

11.4

Figure P11.4 shows a three-story shear frame. The story stiffnesses and floor masses are as indicated. Form the stiffness and the mass matrices of the frame. Obtain an estimate of the fundamental frequency by the Rayleigh method and an assumed vibration shape given by q1T = [1 2 3] Using inverse iteration with a shift, obtain a more precise value of the fundamental frequency and the corresponding mode shape of the building.

598

Dynamics of structures

Figure P11.4 11.5

The stiffness and mass matrices of the three-story building frame of Figure P11.5 are shown on the figure. There is a concern for possible resonant vibrations due to a blower fan mounted at the second-story level. The operating speed of the motor is 250 rpm. Obtain the natural frequency of the building frame that is closest to the blower frequency.

Figure P11.5 11.6

Using Lanczos vectors reduce the eigenvalue problem for the frame of Figure P11.5 to a tridiagonal form.

11.7

From the tridiagonal matrix obtained in Problem 11.6, compute the three natural frequencies of the frame by using the QR method.

11.8

Obtain estimates of the first two frequencies of the frame of Figure P11.5 by using the determinant search method.

11.9

A train made up of 3 cars of weight 60,000 lb each connected by coupling springs of stiffness 15,000 lb/in. are released by a shunting engine and roll freely along a track (Figure P11.9). Determine the mode shapes and frequencies of vibration of the system of cars by sweeping the rigid body mode.

Numerical solution of the eigenproblem

599

Figure P11.9

11.10

Determine the mode shapes and frequencies of the system of cars in Problem 11.9 by adding a fictitious spring to restrain the rigid body mode.

11.11

Solve Problem 11.9 by using vector iteration with shift.

11.12

A small two engined plane in flight is modeled as shown in Figure P11.12. By sweeping off the rigid body modes of the plane obtain a reduced model and hence determine the remaining frequency or frequencies of the plane. Neglect the rotatory inertia of the engine masses and the fuselage.

Figure P11.12 11.13

The stiffness and mass matrices of a discrete parameter system are as given below. 



2

  M=  

    

2 2 2 1



2800  −1200  0 K =   0 0

−1200 2400 −1200 0 0

0 −1200 2000 −800 0

0 0 −800 1200 −400

 0 0  0  −400 400

Using subspace iteration and the following trial vectors, obtain the first two mode shapes and frequencies of the system. Assume that acceptable accuracy has been achieved after three iterations.   0.2 −0.5 0.4 −1.0    0 =  0.6 −0.5 0.8 0.5 1.0 1.0

600 11.14

Dynamics of structures The mass and stiffness properties of a four-story shear frame are shown in Figure P11.14. Using the following trial vectors and subspace iteration obtain the first two mode shapes and frequencies of the frame. 

0.50 1.00  1.19 1.48

 0.50 1.00  −0.05 −1.62

Figure P11.14 11.15

Transform the eigenvalue equation in Problem 11.5 to standard symmetric form by using Equations 11.16 and 11.17. Now reduce the resulting problem to a tridiagonal form by using Householder’s transformation.

11.16

From the tridiagonal matrix obtained in Problem 11.15 compute the three frequencies and mode shapes by using the QR method.

Chapter 12

Forced dynamic response: Multi-degree-of-freedom systems

12.1

INTRODUCTION

The dynamic response of a multi-degree-of-freedom system is governed by the following equations of motion: Mu¨ + Cu˙ + Ku = p(t)

(12.1)

in which p(t) is the vector of time-dependent forcing functions. If the system is undamped, the damping term Cu˙ will be absent from the equations of motion. Equation 12.1 represents N simultaneous equations in the displacement vector u and its time derivatives. These equations can, in general, be solved only by a numerical method of integration. Because such methods of integration are based on certain mathematical idealizations, usually related to the manner in which the accelerations vary over a short period, the results obtained are only approximate. An exact mathematical solution may be possible when the time-varying forces are simple mathematical functions that would permit integration of the equations. Such exact solutions involve a transformation of the equations to a special set of coordinates called normal coordinates. The transformation uncouples the equations of motion so that they are reduced to N independent equations whose mathematical solutions are relatively easily obtained. The method based on normal coordinate transformation, called the mode superposition method, is similar to that described in Chapter 10 for the solution of free-vibration problems of multi-degree-of-freedom systems. Application of the mode superposition method to the solution of forced dynamic response is described in this chapter. Direct integration of the equations of motion by numerical methods is discussed in Chapter 13.

12.2

NORMAL COORDINATE TRANSFORMATION

The application of normal coordinate transformation to obtain the solution to freevibration problems of a multi-degree-of-freedom system was described in Section 10.7. The transformation is given by Equation 10.56, restated here for ease of reference u=

N

n=1

φn yn

(12.2a)

602

Dynamics of structures

or u = y

(12.2b)

where  is the matrix of undamped mode shapes and y is the vector of normal coordinates. On applying this transformation to Equation 12.1 and using the procedure of Section 10.7, we get the following set of transformed equations: M∗ y¨ + C∗ y˙ + K∗ y = p∗

(12.3)

where M∗ = T M C∗ = T C ∗

(12.4)

K =  K T

As explained in Chapter 10, because of the orthogonality property of the mode shapes, both the mass matrix and the stiffness matrix are diagonal. In fact, if the mode shapes have been normalized to be mass orthonormal, M∗ is an identity matrix. Further, if the damping matrix possesses the orthogonality property, C∗ is also diagonal. In such a case, Equation 12.3 reduces to a set of N uncoupled equations. If the nth diagonal term in the transformed mass matrix is denoted by Mn , that in the transformed damping matrix by Cn , and that in the transformed stiffness matrix by Kn , the nth equation in the set of transformed equations becomes Mn y¨ n + Cn y˙ n + Kn yn = pn

(12.5)

where pn = φnT p. For the sake of generality we have retained the term Mn , even though with mass orthonormal modes it will be equal to unity. Equation 12.5 is the equation of motion of a single-degree-of-freedom system. On dividing through by Mn it reduces to y¨ n +

Cn Kn pn y˙ n + yn = Mn Mn Mn

(12.6)

As for a single-degree-of-freedom system, Cn /Mn = 2ξn ωn and Kn /Mn = ωn2 , where ωn is the nth natural frequency and ξn is the damping ratio in the nth mode. Equation 12.6 therefore becomes y¨ n + 2ξn ωn y˙ n + ωn2 yn =

pn Mn

(12.7)

It is of interest to note that in Equation 12.2a, φn yn represents the contribution of the nth mode of vibration to the total response. Further, yn , the nth normal coordinate, is completely determined by solving Equation 12.7, which involves only those parameters that are related to the nth mode. The total response in Equation 12.2 can thus be viewed as a superposition of the responses in individual modes. It is because of this

Forced dynamic response: Multi-degree-of-freedom systems 603

fact that the solution procedure being discussed here is called the mode superposition method. In deriving Equation 12.7, it is only necessary to know the damping ratio ξn , not the damping term Cn , which can only be determined provided that the damping matrix C is specified. As already mentioned, it is easier to assign a reasonable value to ξn than to determine the damping matrix C. It is therefore clearly an advantage of the mode superposition method that it does not require an explicit specification of the damping matrix. It remains to solve the single-degree-of-freedom equation, Equation 12.7, to determine the generalized coordinate yn . Methods of solution of such an equation were discussed in detail in Chapters 5 through 9. Complete solution of Equation 12.7 consists of a free- and a forced-vibration component. The former is given by Equation 5.37, while the latter can be obtained from Duhamel’s integral. The complete solution is thus given by the following equation, which is identical to Equation 7.10:

yn = e

−ξn ωn t

+




y˙ 0n + y0n ξn ωn sin ωDn t + y0n cos ωDn t ωDn  t pn (τ )e−ωξn (t−τ ) sin ωDn (t − τ ) dτ

1 Mn ωDn



(12.8)

0

As noted in Chapter 10, y0n and y˙ 0n in Equation 12.8 are the initial values of yn and y˙ n , respectively, determined from the given initial conditions by using Equations 10.61 and  10.62. The damped frequency ωDn = ωn 1 − ξn2 . When pn (t) is a simple mathematical function of time t, Equation 12.8 can be evaluated exactly. In other cases, some form of numerical integration technique must be used. For a complete solution of the original problem, Equation 12.8 must be solved for each of the N mode shapes. The response in the physical coordinates is obtained by superposing the modal responses by using Equation 12.2. In many practical problems, a major portion of the response is contained in only a few of the mode shapes, usually those corresponding to the lowest frequencies. In such cases, modal response need be obtained for only the first few of the N modes, and therefore only such mode shapes and frequencies that are required need to be calculated. For large problems, this may mean a significant saving in the computational time. In the mode superposition method described above, the undamped mode shapes were used to carry out a transformation of coordinates. The transformation uncouples the equations of motion provided that the damping matrix C satisfies the orthogonality condition. In a general case, where C does not satisfy such a condition, uncoupling can still be achieved provided that complex eigenvectors of the damped system are used in the transformation. The procedure is similar to that described in Section 10.8 for the solution of a damped free-vibration problem. However, as noted in that section, the determination of complex eigenvalues and eigenvectors involves large computational effort. Besides, the representation of damping in the system by a damping matrix is only an idealization. It is therefore more expedient and equally reasonable to use the

604

Dynamics of structures

undamped mode shapes in the transformation and to select a C that would become diagonal under such a transformation.

12.3

SUMMARY OF MODE SUPERPOSITION METHOD

We can now summarize the steps involved in mode superposition analysis of the forced vibration of a multi-degree-of-freedom system. 1

From the specified physical characteristics of the system, obtain the mass, stiffness, and damping matrices. Formulate the equations of motion using the physical property matrices and the given force vector Mu¨ + Cu˙ + Ku = p

2

(12.9)

For the mode superposition method to be successful, the damping matrix C must possess the orthogonality property. In most cases, the physical damping characteristics are not known. Damping is then specified at the modal level and the damping matrix C need not be determined. Obtain the free-vibration mode shapes and frequencies of the undamped system by solving the following linearized eigenvalue problem by one of the methods described in Chapter 11: Mu¨ + Ku = 0

(12.10a)

Kφ = ω2 Mφ

(12.10b)

or

3

4

in which ω is a frequency and φ is the corresponding mode shape. As stated earlier, in the mode superposition method, it is not necessary to use all N mode shapes, N being the number of degrees of freedom. Reasonable accuracy can be achieved by using only a few modes. How many modes need to be used in a specific case is a question that we deal with more fully in Chapter 13; for the present discussion let us assume that the first M modes will give reasonable accuracy in the determination of response. It will then be necessary to determine only the first M mode shapes and frequencies. For each of the M mode shapes, determine the modal mass Mn , the modal damping Cn and the modal force pn , given by Mn = φnT Mφn

(12.11a)

Cn = φnT Cφn

(12.11b)

pn =

(12.11c)

φnT p

Note that since the mode shapes used in Equations 12.11 are mass orthonormal Mn is equal to unity. Also, if damping is to be specified at the mode level, it would

Forced dynamic response: Multi-degree-of-freedom systems 605

5

be given in the form of a damping ratio ξn and the damping constant would then be given by Cn = 2ξn ωn Mn . The equation of motion corresponding to the nth mode now becomes y¨ n +

Cn Kn pn y˙ + y= Mn Mn Mn

(12.12a)

or y¨ n + 2ξn ωn y˙ n + ωn2 yn = 6

pn Mn

(12.12b)

From the given initial conditions in the physical coordinates, obtain the values of y0 and y˙ 0 , where y0 is the vector of modal displacements at time t = 0, and y˙ 0 is the vector of the time derivatives of the modal displacements, again at time t = 0. It is possible to obtain these directly from Equation 12.2b as follows y0 = −1 u0

(12.13a)

−1

y˙ 0 =  v0

(12.13b)

It is not, however, necessary to use Equations 12.13, which involve the inverse of ; instead, y0 and y˙ 0 can be obtained more simply by using the mass orthonormality property of the mode shapes as in Equations 10.61 and 10.62 y0 = T Mu0

(12.14a)

y˙ 0 =  Mv0

(12.14b)

T

7

Obtain solutions to the M single-degree-of-freedom equations given by Equation 12.12 for n = 1 to M. The Duhamel integral solution is given by


y˙ 0n + y0n ξn ωn sin ωDn t + y0n cos ωDn t ωDn  t 1 + pn (τ )e−ωn ξn (t−τ ) sin ωDn (t − τ )dτ Mn ωDn 0

yn = e−ξn ωn t

8



(12.15)

The integral in Equation 12.15 can be evaluated mathematically if pn (τ ) is a simple function; otherwise, a method of numerical integration must be used. The response in physical coordinates can now be obtained by superposing the modal responses

u=

M

n=1

φ n yn

(12.16)

606

Dynamics of structures

9

If desired, the spring forces can be obtained by using the relationship fS = Ku =

M

Kφn yn

(12.17)

n=1

Since the mode shapes φn satisfy the relationship Kφn = ωn2 Mφn , Equation 12.17 can be expressed as fS =

M

ωn2 Mφn yn

(12.18)

n=1

Both Equations 12.17 and Equation12.18 show that the spring forces are also obtained as a superposition of the modal contributions. Example 12.1 Derive the equations of motion for the system shown in Figure E12.1 and calculate the mode shapes and frequencies. Obtain the response of the system when it is subjected to a suddenly applied constant force F along coordinate 3. Given m = 1, k = 1, and F = 1.

Figure E12.1 Three-degree-of-freedom system, forced vibrations.

Solution We use an energy method to derive the equations of motion. The kinetic energy, T, of the system is given by T=

1 1 1 1 mu˙ 21 + 2mu˙ 22 + I θ˙ 2 + 2mu˙ 23 2 2 2 2

(a)

where θ˙ is the angular velocity and I is the mass moment of inertia of the roller. Since the roller is assumed to roll without slipping, u2 − u1 = aθ

(b)

Substituting θ˙ from Equation b and the value I = ma2 in Equation a, we get T=

1 1 mu˙ 21 + mu˙ 22 + m(u˙ 2 − u˙ 1 )2 + mu˙ 23 2 2

(c)

Forced dynamic response: Multi-degree-of-freedom systems 607 The potential energy stored in the springs is given by V=

1 1 2 1 ku + 2k(u2 − u1 )2 + 2k(u3 − u1 )2 2 1 2 2

(d)

Using Lagrange’s equations (Equation 4.91), the equations of motion are obtained as follows 2mu¨ 1 − mu¨ 2 + 5ku1 − 2ku2 − 2ku3 = 0 −mu¨ 1 + 3mu¨ 2 − 2ku1 + 2ku2 = 0

(e)

2mu¨ 3 − 2ku1 + 2ku3 − F = 0 In matrix form, Equations e can be expressed as 

−m

2m

 −m 0

3m 0

   5k −2k u¨ 1    0  u¨ 2  + −2k 2k u¨ 3 2m −2k 0 0

    0 u1     0   u2  =  0  u3 2k F

−2k

(f)

or Mu¨ + Ku = p The eigenvalue problem is given by Equation f with the right-hand side equal to zero. Its solution can be obtained from one of the standard method of eigensolution described in Chapter 11. The details of calculation are not shown, but the resulting eigenvalues and eigenvectors are as follows λ1 = 0.1703

ω1 = 0.4127

λ2 = 0.7663

ω2 = 0.8754

λ3 = 3.063

ω3 = 1.7503



(g)



1.0000

  q1 = 1.2289 1.2054 



1.0000

  q2 = −4.1268

(h)

4.2792 

 1.0000   q3 =  0.1479 −0.4847 The modal matrix is given by Q = [q1

q2

q3 ]

(i)

608

Dynamics of structures

A normal coordinate transformation is now carried out using the modal matrix Q, so that u = Qy ˜ = QT MQ M ˜ = QT KQ K

(j)

p˜ = QT p and the transformed equations are ˜ y + Ky ˜ = p˜ M¨

(k)

On carrying out the computations, we will find that the set of equations given by Equation k are uncoupled. The uncoupled equations are 6.9788¨y1 + 1.1892y1 = 1.2054F 97.969¨y2 + 75.073y2 = 4.2792F

(l)

2.2396¨y3 + 6.8603y3 = −0.4847F The solutions to Equations l can be obtained from Equation 7.14 with the damping ratio ξ = 0 1.2054F (1 − cos 0.4127t) 1.1892 4.2792F (1 − cos 0.8754t) y2 = 75.073 −0.4847F (1 − cos 1.7503t) y3 = 6.8603 y1 =

(m)

The response in the physical coordinates is now obtained by using the transformation u = Qy, giving u1 = 0.9999 − 1.0136 cos 0.4127t − 0.0570 cos 0.8754t + 0.0707 cos 1.7503t u2 = 0.9999 − 1.2456 cos 0.4127t + 0.2352 cos 0.8754t + 0.0105 cos 1.7503

(n)

u3 = 1.500 − 1.2218 cos 0.4127t − 0.2439 cos 0.8754t − 0.0343 cos 1.7503t

12.4

COMPLEX FREQUENCY RESPONSE

As in the case of a single-degree-of-freedom system, the response of a multi-degree-offreedom system to a harmonic excitation can conveniently be expressed in the notation of complex algebra. Assume that the forcing function is given by p = feit

(12.19)

Forced dynamic response: Multi-degree-of-freedom systems 609

where f represents the spatial variation of force and is independent of time. The equation of motion of a damped multi-degree-of freedom system subjected to the exciting force given by Equation 12.19 can be expressed as Mu¨ + Cu˙ + Ku = feit

(12.20)

The steady-state solution to Equation 12.20 is u = Ueit

(12.21)

Substitution of Equation 12.21 in Equation 12.20 gives [−2 M + iC + K]Ueit = feit

(12.22)

Equation 12.22 can be solved for U to yield U = [−2 M + iC + K]−1 f = Rd f

(12.23)

As in the case of a single-degree-of-freedom system Rd is referred to as the receptance matrix. It can be obtained by taking the inverse of matrix [−2 M + iC + K]. This, however, involves considerable amount of computations, particularly because the matrix is complex-valued. For a system with proportional damping a simpler and more instructive procedure is to obtain a solution by transforming the equations of motion to normal coordinates. The transformed equations are obtained from either Equation 12.5 or 12.7. y¨ n + 2ξn ωn y˙ + ω2 yn = φnT feit

n = 1, 2, . . . , N

(12.24)

The steady-state solution of Equation 12.24 is given by: yn = Yn eit

(12.25)

where Yn =

(ωn2

−

φn f + 2iξn ωn 

2 )

(12.26)

The solution in physical coordinates is given by: U=

N

φn Yn

n=1

Rd f =

n

φn φnT f (ωn2 − 2 ) + 2iξn ωn 

(12.27)

610

Dynamics of structures

The receptance matrix Rd is an N × N matrix, in which the term (Rd )lk represents the displacement at coordinate l produced by a unit harmonic load applied at coordinate k. For l = k, (Rd )kk is referred to as a point receptance. When l = k, (Rd )lk is called cross receptance. The velocity response to harmonic excitation is obtained by differentiating Equation 12.21 u˙ = iUeit = iRd feit = Rv feit

(12.28)

where Rv = iRd is referred to as the mobility matrix. The acceleration response is given by u¨ = −2 Rd feit = Ra feit

(12.29)

in which Ra = −2 Rd is called inertance matrix. It is instructive to examine an individual element of matrix Rd , given by (Rd )lk =

n

(ωn2

φln φkn − 2 ) + 2iξn ωn 

(12.30)

in which φln is the lth term of φn , φkn is the kth term of φn and summation extends over all N modes. It is evident from Equation 12.30 that (Rd )lk = (Rd )kl and matrix Rd is symmetric. Also, an individual term in the summation depends only on the parameters related to a single mode and, of course, the excitation frequency. Graphical representation of the complex frequency response function (FRF) given by Equation 12.30 may consist of a plot of the amplitude of FRF versus the excitation frequency  and a plot of the phase angel versus . Alternatively, a Nyquist graph, in which the imaginary part of the FRF is plotted against the real part, may be used. As an illustration, consider the undamped two-degree-of-freedom system shown in Figure 12.1. Assume that k2 = 0.5k1 and m2 = 0.5m1 . The two mode shapes and

Figure 12.1 Harmonic force excitation of a two-degree-of-freedom system.

Forced dynamic response: Multi-degree-of-freedom systems 611

frequencies of the system are % % k1 2k1 ω2 = ω1 = 2m1 m1 1  φ1 =

√

3m1 √2 3m1

φ2 =

 √ 1 1.5m1 1 − √1.5m 1

The elements of the receptance matrix are obtained on substituting appropriate values in Equation 12.30. 1/3k1 1/1.5k1 ∗ 2 + 1/2 − (/ω1 ) 2 − (/ω1∗ )2 2/3k1 2/3k1 (Rd )12 = (Rd )21 = ∗ 2 − 1/2 − (/ω1 ) 2 − (/ω1∗ )2 4/3k1 2/3k1 + (Rd )22 = 1/2 − (/ω1∗ )2 2 − (/ω1∗ )2

(Rd )11 =

(12.31a) (12.31b) (12.31c)

 where ω1∗ = k1 /m1 . The amplitudes of the two terms in the point receptance given by Equation 12.31a are plotted in Figure 12.2. Also shown in that figure is the

Figure 12.2 Frequency response functions for a two-degree-of-freedom system, point receptance plots.

612

Dynamics of structures

resultant value (Rd )11 . It may be noted that the amplitudes of the individual terms in Equation 12.31a represent the absolute values of these terms; the sign being reflected in the phase angle, which is either 0◦ or 180◦ . However, in taking the sum of the two terms to get their resultant, account must be taken of the signs of the individual terms. Several interesting points may be noted by observing the plots shown in Figure 12.2. When the excitation frequency is close to one of the modal frequencies, the term related to that mode dominates in the expression for FRF. In general it can be stated that for well-separated modes the following provides a reasonable estimate of the frequency response.

(Rd )lk ≈

φlm φkm 2 − 2 ) + 2ξ ω  (ωm m m

for  ≈ ωm

(12.32)

When  coincides with one of the natural frequencies, the response for the undamped system becomes infinite. This phenomenon is referred to as resonance.

Figure 12.3 Frequency response functions for a two-degree-of-freedom system, cross-receptance plots.

Forced dynamic response: Multi-degree-of-freedom systems 613

For excitation frequencies lying between two consecutive natural frequencies, that is ω1 <  < ω2 , the two terms in the point receptance, for example those in 2 Equation 12.31a, are of opposite sign. This is because the numerator φ1n is always 2 2 positive; on the other hand, while the denominator ω1 −  is negative, ω22 − 2 is positive. Consequently, at some value of  between ω1 and ω2 the two terms cancel each other and the resultant response is zero. In the present case this happens when /ω1∗ = 1. A point in the FRF where the response becomes very small is called antiresonance. Resonances are also evident in the cross-receptance plot (Rd )12 shown in Figure 12.3. However, antiresonance does not exist between ω1 and ω2 , although a minimum still exists. In the present case, the two terms in the expression for (Rd )12 are both positive and do not therefore cancel each other. Example 12.2 Assume that viscous damping exists in the two-degree-of-freedom system shown in Figure 2.1 and is represented by damping ratios ξ1 = ξ2 = 0.1 in each of the two modes. Also, k2 = 0.5k1 and m2 = 0.5m1 . Plot the amplitude and phase angle against the exciting frequency for the receptance (Rd )11 . Also plot the Nyquist diagrams for both (Rd )11 and (Rd )12 .

Solution The undamped frequencies and mode shapes of the system were derived earlier and are given by % ω1 =  φ1 =

% k1 2m1 √1

ω2 = 

 3m1  √2 3m1

2k1 m1 

φ2 =

√



1  1.5m1  −√ 1 1.5m1

The receptances are obtained by substituting the above values in Equation 12.30

(Rd )11 =

1 2 + √ √ ¯ ¯ ¯ 2 + i 2ξ1 ) ¯ 2 + i2 2ξ2 ) 3k1 (0.5 −  3k1 (2 − 

(a)

(Rd )12 =

2 2 − √ √ ¯ ¯ ¯ 2 + i 2ξ1 ) ¯ 2 + i2 2ξ2 ) 3k1 (0.5 −  3k1 (2 − 

(b)

 ¯ = /ω1∗ and ω1∗ = k1 /m1 where  The amplitude and phase angle for (Rd )11 are plotted in Figure E12.2a. Both resonance and antiresonance exist in the amplitude plot, however, damping has caused a smoothening of the curves in each case. The Nyquist diagram for (Rd )11 is obtained by plotting the imaginary part against the real part in Equation a and is shown in Figure E12.2b. The Nyquist plot for (Rd )12 obtained from Equation b is shown in Figure 12.2c.

614

Dynamics of structures

Figure E12.2 Frequency response of a damped two-degree-of-freedom system; (a) point receptance and phase angle; (b) Nyquist plot for point receptance; (c) Nyquist plot for cross-receptance.

Forced dynamic response: Multi-degree-of-freedom systems 615

Figure E12.2 Continued.

12.5 VIBRATION ABSORBERS Vibration absorbers are devices that are used to kill or reduce unwanted vibrations in mechanical or structural systems. Although the detailed design of a vibration absorber must consider many factors, the conceptual background to the design is quite straightforward and relies on the phenomenon of antiresonance discussed in the preceeding section. Referring to Figure 12.1, let the primary system be represented by a singledegree-of-freedom system having a mass m1 and spring stiffness k1 . Vibrations in the system are caused by a harmonic force p0 eit acting on the mass m1 . To reduce these vibrations a mass m2 is attached to the primary system through a spring of stiffness k2 . The response of the combined system can be obtained from Equations 12.21 and 12.27. It can be shown that the displacement amplitudes U1 and U2 are given by k2 − m 2 2 D k2 U2 = p0 D

U 1 = p0

(12.33)

616

Dynamics of structures

where D is obtained from D = (k1 + k2 − m1 2 )(k2 − m2 2 ) − k22

(12.34)

Equation 12.33 shows that the vibrations of the primary system will be completely killed if the characteristics of the vibration absorbers are selected so that its frequency is equal to the operating frequency , that is % ω2∗ =

k2 = m2

(12.35)

In fact the FRF (Rd )11 reaches an antiresonance when this condition is satisfied, and for operating frequencies in the vicinity of antiresonance, vibrations of the primary system are reduced quite considerably. Because the characteristics of the vibration absorber are matched to a specific requirement, it is also referred to as a tuned mass damper. Obviously, a tuned mass damper is only effective if the operating frequencies are known to lie within a narrow range. The size and mass required for the absorber may also pose practical difficulty in its design,  particularly when the primary structure is massive. At antiresonance, that is when k2 /m2 = , the denominator D in Equation 12.33 becomes −k22 , so that the displacement amplitude for mass m2 becomes U2 = −

p0 k2

(12.36)

Equation 12.36 shows that the force exerted by the vibration absorber exactly balances the applied force. For a massive structure p0 is expected to be large, and to keep the displacement of absorber mass within a reasonable bound, k2 must be large. At the same time a large k2 should be accompanied by a large value of m2 , so as to satisfy Equation 12.35. As an example of the effectiveness of a tuned mass damper consider the case when m2 = 0.1m1 and k2 = 0.1k1 . The FRF for the response of primary systemafter the attachment of tuned mass damper is shown in Figure 12.4, in which ω1∗ = (k1 /m1 ) is the uncoupled frequency of the primary system. Antiresonance is achieved when  = ω2∗ . Note that in the present example ω2∗ = ω1∗ . Suppose that the design calls for the dynamic displacements not to exceed the static displacement under p0 . The range of operating frequencies over which the tuned mass damper will be effective are then as shown in Figure 12.4.

12.6

EFFECT OF SUPPORT EXCITATION

In many situations, a system may be made to vibrate by the motion of supports to which it is attached. Support-induced vibrations cause deformations and stresses in the structural or mechanical system, just as vibrations due to applied dynamic forces do. Obviously, such deformations and stresses must be taken into account in the design of the system.

Forced dynamic response: Multi-degree-of-freedom systems 617

Figure 12.4 Frequency response function for a system equipped with vibration absorber.

Many different examples can be cited of systems excited by support movement. Dynamic vibrations of an instrument attached to a moving frame is one such example. Earthquake-induced vibrations of a building is another example. In calculating the dynamic response of a system excited by support motions, we will, as usual, need the mass, stiffness, and damping matrices of the system. There are, however, no external forces except those acting at the attachments of the system to the supports. For a general treatment of the response, we include in our coordinates the displacement degrees of freedom at the supports as well. The mass matrix of the system can then be partitioned as follows

Maa M= Mba

Mab Mbb

 (12.37)

where suffix a denotes superstructure degrees of freedom and suffix b denotes support degrees of freedom. Matrices Mab and Mba denote coupling mass matrices between the superstructure and the supports. Note that in a lumped mass idealization, both Mab and Mba will be null matrices while Maa and Mbb will be diagonal. Coupling terms will exist, for example, when a consistent mass formulation is used (Section 3.6.2).

618

Dynamics of structures

The stiffness and damping matrices are partitioned in a similar manner.

Kaa K= Kba

Caa

C=

Cba

Kab

 (12.38)

Kbb Cab Cbb

 (12.39)

The force vector is given by f=

 0

(12.40)

fb

Note that there are no external forces along the superstructure degrees of freedom. We now denote the displacements along the superstructure degrees of freedom by uat and those along the base degrees of freedom by ub . The superscript t on the superstructure displacements signifies that these displacements are absolute and is used to differentiate the absolute displacements from the relative displacements, which we use later in the formulation. With the notations described above, the equations of motion can be written as Maa Mba

Mab Mbb

  u¨ at u¨ b

+

Caa

Cab

Cba

Cbb

  u˙ at u˙ b

+

Kaa

Kab

Kba

Kbb

  uat ub

=

 0 fb

(12.41)

Equations 12.41 can be expressed in expanded form as Maa u¨ at + Caa u˙ at + Kaa uat + Mab u¨ b + Cab u˙ b + Kab ub = 0 Mbb u¨ b + Cbb u˙ b + Kbb ub +

Mba u¨ at

+

Cba u˙ at

+

Kba uat

= fb

(12.42a) (12.42b)

In Equations 12.42, the mass, damping, and stiffness matrices can be determined from the known physical characteristics of the system, while the support motions ub , u˙ b , and u¨ b must be specified. The unknowns to be determined are the displacements along the superstructure degrees of freedom, ua , and the support forces, fb . To reduce Equations 12.42 to a more useful form, let us examine the response of the system when the support displacements ub are applied in a static manner. This is possible provided we assume that ub can be expressed as the product of a vector corresponding to the spatial degrees of freedom multiplied by a single time function, so that ub = u˜ b f (t)

(12.43)

where f (t) denotes a function of time. Static application of ub means that the time variation is absent and the applied displacement vector is therefore equal to u˜ b . The response caused by the application of

Forced dynamic response: Multi-degree-of-freedom systems 619

u˜ b can be obtained from Equation 12.41 by noting that the velocities and accelerations will be zero. Thus    0 Kaa Kab u˜ as = s (12.44) ˜ Kba Kbb u˜ b f b

in which u˜ as represents displacement of the superstructure and f˜bs represents support forces, both obtained from a static application of u˜ b . Equation 12.44 gives Kaa u˜ as + Kab u˜ b = 0

(12.45a)

or −1 Kab u˜ b u˜ as = −Kaa

= Ru˜ b

(12.45b)

where −1 Kab R = −Kaa

(12.46)

On multiplying both sides of Equation 12.45 by f (t), we get u˜ as f (t) = Ru˜ b f (t)

(12.47a)

uas = Rub

(12.47b)

or

where uas = u˜ as f (t) can be defined as a pseudostatic response. We now express the total displacements uat as a superposition of the pseudostatic displacements uas and relative displacements ua , so that uat = uas + ua = Rub + ua

(12.48)

Substitution in Equation 12.42a gives Maa u¨ a + Caa u˙ a + Kaa ua + Maa Ru¨ b + Mab u¨ b + Caa Ru˙ b + Cab u˙ b + Kaa Rub + Kab ub = 0

(12.49)

Now because (Kaa Rub + Kab ub ) = 0 according to Equation 12.45a, Equation 12.49 reduces to Maa u¨ a + Caa u˙ a + Kaa ua = −(Maa R + Mab )u¨ b − (Caa R + Cab )u˙ b

(12.50)

620

Dynamics of structures

Equation 12.50, written in terms of the relative displacements ua , is entirely equivalent to Equation 12.1, with the forcing function p(t) being a function of the base velocity u˙ b and base acceleration u¨ b . It can therefore be solved by application of the mode superposition method using the mode shapes of a fixed-base undamped system in exactly the same manner as for a general applied force vector. Once the displacement vector ua has been determined by solving Equations 12.50, the spring forces can be obtained in the usual manner. Thus, from Equation 12.41, fS = Kaa uat + Kab ub = Kaa ua + (Kaa uas + Kab ub )

(12.51)

= Kaa ua in which Kaa uas + Kab ub has been set as equal to zero because of the relationship in Equation 12.45. The elastic forces along the support degree of freedom are obtained from fSb = Kba uat + Kbb ub

(12.52a)

fSb = Kba ua + (Kba uas + Kbb ub )

(12.52b)

or

The terms within parentheses in Equation 12.52b are together equal to f˜bs f (t), as seen from Equation 12.44, and thus represent the pseudostatic elastic forces caused by the displacements of the supports. On substitution for uas from Equation 12.45, Equation 12.52b becomes −1 fSb = Kba ua + (Kbb − Kba Kaa Kab )ub

(12.53)

Equation 12.50 can be considerably simplified for certain special cases. For example, if the damping matrix is proportional to the stiffness matrix, the term Caa R + Cab will be zero. This can easily be verified by setting Caa = βKaa and Cab = βKab , where β is a constant of proportionality, and then substituting for R from Equation 12.46. When the damping matrix is not proportional to stiffness matrix, the contribution of the damping terms to the forcing function will not be zero. In practice, however, this contribution is relatively small and can often be neglected. A major simplification results when the motion is identical for all the supports. In such a case, ub can be expressed as ub = 1ug

(12.54)

where ug is the common support motion and 1 is a vector of p elements each equal to 1, p being the number of base degrees of freedom. Equation 12.45b now gives −1 uas = Kaa Kab 1ug

= rug

(12.55)

Forced dynamic response: Multi-degree-of-freedom systems 621

where −1 r = −Kaa Kab 1

(12.56)

It should be noted that for identical support motions, the system moves as a rigid body; vector r need not therefore be evaluated from Equation 12.56 but can be obtained simply by kinematic considerations. For identical support motion and a negligible damping contribution to the forcing function, Equation 12.50 becomes Maa u¨ a + Caa u˙ a + Kaa u¨ a = −Maa ru¨ g − Mab 1u¨ g

(12.57)

Further, because for a rigid-body motion, f˜bs in Equation 12.44 should be zero, Equation 12.52b for elastic support forces reduces to fSb = Kba ua

(12.58)

Finally, for a mass idealization in which the coupling terms Mab are absent, Equation 12.57 simplifies to Mu¨ + Cu˙ + Ku = −Mru¨ g

(12.59)

where, for simplicity, we have dropped the subscript a. Application of the mode superposition method to the solution of Equation 12.58 will provide N uncoupled equations of motion, where N is the number of superstructure degrees of freedom and it is assumed that the damping matrix possesses the orthogonality property. The nth uncoupled equation will be y¨ n + 2ξn ωn y˙ n + ωn2 yn = −

φnT Mr u¨ g Mn

(12.60)

The uncoupled equations can be solved by standard methods of analysis. Response in the physical coordinates is then obtained by a superposition of the modal responses. Example 12.3 Formulate the equations of motion for support excitation of the system shown in Figure E12.3. Obtain expressions for the motion of masses 1 and 2 when the support excitation is given by (a) u¨ g =

(b) u¨ g =

 1 2  1 1

f (t)

f (t)

The mass, damping, and stiffness parameters are: m1 = 1, m2 = 2, m3 = 12 , m4 = 12 , c1 = 3.04, c2 = 6.08, c3 = 3.04, k1 = 600, k2 = 1200, and k3 = 600.

622

Dynamics of structures

Figure E12.3 Response of a multi-degree-of-freedom system to support excitation.

Solution The mass, damping, and stiffness matrices are obtained quite easily and are given by 



m1

  M= 

   

m2 m3

(a)

m4 

c1 + c2

 −c 2  C=  −c1 0

 k1 + k2  −k 2  K=  −k1 0

−c2

−c1

c2 + c 3

0

0

c1

−c3

0

−k2

−k1

k2 + k 3

0

0

k1

−k3

0



0

−c3    0 

(b)

c3 0



−k3    0 

(c)

k3

On substituting the values of mass, stiffness, and damping parameters, we get 

1

0  M= 0 0

0

2

0

0

1 2

0   0

0

0

1 2



0



0

9.12 −6.08  C= −3.04

−6.08

0.00

−3.04

1800



−1200  K=  −600 0



−3.04

0.00

0.00

3.04

−1200 −600

0

0.00 −3.04   3.04 0.00

9.12 0.00

1800 0 −600



0 −600   600 0 0

600

Forced dynamic response: Multi-degree-of-freedom systems 623 (a) The psuedostatic displacement is given by uas = −Rub

(d)

where −1 R = −Kaa Kab

Substituting the submatrices

 −1200

Kaa =

1800 −1200

1800

Kab =



−600

0

0

−600

in Equation d, we get 1 3 R= 5 2



2 3

Because damping matrix is proportional to the stiffness matrix, we have Caa u˙ as + Cab u˙ b = 0 and the equation of motion corresponding to the superstructure degrees of freedom becomes Maa u¨ a + Caa u˙ a + Kaa ua = −Maa Ru¨ b

(e)

or

1 0

    0 9.12 −6.08 1800 −1200 1 7 u¨ a + f (t) u˙ a + ua = − 5 16 2 −6.08 9.12 −1200 1800

To solve Equation e by the mode superposition method, we need the fixed-base undamped mode shapes of the system. These are obtained by solving the following free-vibration equations: Maa u¨ a + Kaa ua = 0

(f)

or



1

0

0

2

u¨ a +



 1800 −1200 −1200

1800

ua = 0

The eigenvalue problem of Equation f can be solved by any one of the standard methods described in Chapter 11. For a problem of small size such as this, the characteristic equation may be solved directly for the two eigenvalues. Substitution of the eigenvalues, in turn, in Equation f will give

624

Dynamics of structures

the corresponding eigenvectors. The following results are obtained: λ1 = 389.5

ω1 = 19.74 rad/s

λ2 = 2310.5 ω2 = 48.07 rad/s  1 q1 = 1.175 q2 =

1



−0.425

The transformed matrices are obtained next: ˜ = QT MQ M  1 1.175 1 = 1 −0.425 =

0

0

1.361

0

2

1

1.175

−0.425



1



3.761

˜ = QT CQ C 7.424 = 0





15.936

˜ = QT KQ K  1465.2 0 = 0 3145.2 Also, p˜ = qT p   5.16 f (t) =− 0.04 The two uncoupled equations of motion are thus 3.761¨y1 + 7.424˙y1 + 1465.2y1 = −5.16f (t) 1.362¨y2 + 15.936˙y2 + 3145.2y2 = −0.04f (t) To obtain the damping ratios in the first mode, we note that 2ξ1 ω1 m1 = c1 Substituting ω1 = 19.74, m1 = 3.761, and c1 = 7.424, we get ξ1 = 0.05

(g)

Forced dynamic response: Multi-degree-of-freedom systems 625 In a similar manner, for the second mode 2ξ2 ω2 m2 = c2

(h)

Since ω2 = 48.07, m2 = 1.362, and c2 = 15.931, Equation h gives ξ2 = 0.1217 The equations of motion can now be expressed in the following alternative form y¨ 1 + 2ξ1 ω1 y˙ 1 + ω12 =

k1 = −1.372f (t) m1

(i)

y¨ 2 + 2ξ2 ω2 y˙ 2 + ω22 =

k2 = −0.0294f (t) m2

(j)

The damped frequencies are given by * ωD1 = ω1 1 − ξ12 = 19.72 * ωD2 = ω2 1 − ξ22 = 47.71 The solutions to the equations of motion can be expressed in terms of Duhamel’s integrals. For the first mode  1.372 t y1 = − f (τ )e−19.74×0.05(t−τ ) sin 19.72(t − τ )dτ 19.72 0  t = −0.07 f (τ )e−0.987(t−τ ) sin 19.72(t − τ )dτ 0

In a similar manner, for the second mode  y2 = −0.000616

t

f (τ )e−5.850(t−τ ) sin 47.71(t − τ )dτ

0

The response in the physical coordinates is given by       u1 1 1 y y1 + = −0.425 2 1.175 u2 (b) In this case, both supports undergo identical motion. Hence the pseudostatic displacements of the two masses can be obtained by rigid-body kinetics. Thus uas = rug  1 = ug 1

(k)

where r is the vector of pseudostatic displacements due to a unit displacement of supports. The equations of motion are given by Maa u¨ a + Caa u˙ a + Kaa ua = −Maa ru¨ g

(l)

626

Dynamics of structures

or





1

0

0

2

u¨ a +





9.12 −6.08 −6.08

9.12

u˙ a +

1800 −1200 −1200

1800

 ua = −

 1 f (t) 2

Of course, direct application of Equation e will lead to the same set of equations. Thus the right-hand side of Equation e is obtained from 1 1 −Maa Ru¨ b = − 5

 2



=−

  2 1

2

3

1

f (t)

  1

1

=−

3

2  1 2

1

f (t)

f (t)

Obviously, the use of rigid-body kinetics makes the problem much simpler. The application of mode superposition method follows lines similar to that in part (a). The frequencies, mode, shapes and damping ratios are the same. The transformed force vector is given by p˜ = −



3.35 0.15

f (t)

The modal displacements are obtained as  t 3.35 f (τ )e−0.987(t−τ ) sin 19.72(t − τ )dτ 19.72 × 3.761 0  t = −0.0452 f (τ )e−0.987(t−τ ) sin 19.72(t − τ )dτ

y1 = −

0

and  y2 = −0.00231

t

f (τ )e−5.850(t−τ ) sin 47.71(t − τ )dτ

0

12.7

FORCED VIBRATION OF UNRESTRAINED SYSTEM

The mode superposition method of analysis applies without modification to the analysis of the forced-vibration response of an unrestrained system provided that the rigid-body modes are included in the transformation. The resultant response obtained by the method automatically includes a rigid-body component of motion. A special method must, however, be used in evaluating the mode shapes. Several such methods were described in Section 11.7, and any one of them may be used as appropriate.

Forced dynamic response: Multi-degree-of-freedom systems 627

Example 12.4 The three-mass system shown in Figure E12.4a is suspended by a rigid wire. Describe the motion of the system in free fall after the wire suddenly snaps. Given m1 = m2 = m3 = m k1 = 2k k2 = k

Figure E12.4 Three-mass suspended system released from rest.

Solution When the system is at rest, mass 2 is displaced downward from the unstretched position of the spring by a distance m2 + m3 mg g= k1 k

(a)

where g is the acceleration due to gravity. In a similar manner, mass 3 is displaced downwards by a distance m2 + m3 m3 g 2mg g+ = k1 k2 k

(b)

We now solve for the motion of the released system shown in Figure E12.4b. The system is clearly unrestrained. The displacements calculated in Equations a and b can be treated as the

628

Dynamics of structures

initial displacements of the system. In addition, each mass of the system is subjected to a force equal to mg. The mass and stiffness matrices are given by 



m

 M=



 

m

(c)

m −2k

2k

 K = −2k

3k −k

0

0



 −k k

(d)

The system has one rigid-body mode, given by   1   q1 = 1 1

(e)

the corresponding eigenvalue being λ1 = 0. Each of the remaining two modes will satisfy the constraint equation q1T Mu˜ = 0 or u˜ 1 + u˜ 2 + u˜ 3 = 0

(f)

Equation f can be expressed in the form of a coordinate transformation    −1 u˜ 1    u˜ 2  =  1 u˜ 3 0

 −1   u2 0 u3 1

(g)

or as u˜ = Tu

(h)

where 

−1

 T= 1 0

 −1  0 1

(i)

The mass and stiffness matrices in the transformed space are given by  ˜ = TT MT = 2m M m ˜ = TT KT = K



m m

(j)



9k

3k

3k

3k

(k)

Forced dynamic response: Multi-degree-of-freedom systems 629 The reduced eigenvalue problem becomes ˜ c = λMu ˜ c Ku The eigenvalues and eigenvectors in the reduced space are easily obtained and are given by λ2 = λ˜ 1 = 1.268

k m

k λ3 = λ˜ 2 = 4.732 m  1 q˜ 1 = −3.732 q˜ 2 =

) ω2 = 1.128 ) ω3 = 2.175

k m

(l)

k m



1 −0.268

The coordinate transformation, Equation h, is used to obtain the complete mode shapes giving  2.732 q2 = Tq˜ 1 =  1  −3.732 

(m)

 −0.732 q3 = Tq˜ 2 =  1  −0.268 

On combining q2 and q3 with the rigid body-mode, the modal matrix Q is obtained as 

1

 Q = 1 1

 −0.732  1 1  −3.732 −0.268 2.732

(n)

The equations of motion for the original system are 

1

 m

1

   2 u¨ 1    u ¨ −2 + k    2 u¨ 3 0 1

−2 3 −1

    1 u1     −1 u2  = mg 1 u3 1 1 0

(o)

with the initial conditions u10 = 0 mg u20 = k 2mg u30 = k

(p)

630

Dynamics of structures

These equations are now transformed to normal coordinates using the relationships u = Qy ˜ = QT MQ M

(q)

˜ = QT KQ K p˜ = QT p The transformed equations become 

3 m0 0

   y¨ 1 0 0 0  y¨ 2  + k0 0 y¨ 3 1.608

0 22.39 0

0 28.39 0

    y1 0 3mg 0   y2  =  0  7.61 0 y3

(r)

The transformation has uncoupled the equations of motion and the uncoupled equations are y¨ 1 = g k y2 = 0 m k y¨ 3 + 4.732 y3 = 0 m

y¨ 2 + 1.268

(s)

Before Equations s can be solved, we need the initial conditions in the normal coordinates. These are obtained by using the relationship Qy0 = u0

(t)

Premultiplication of both sides of Equation t by QT M gives QT MQy0 = QT Mu0

(u)

˜ 0 = QT Mu0 My Substitution for Q, M, and u0 gives 

3

 0 0

0 22.39 0

0



y10





3



    mg 0  y20  = −6.464 k y30 1.608 0.464

or y10 =

mg k

mg k mg = 0.2886 k

y20 = −0.2886 y30

(v)

Forced dynamic response: Multi-degree-of-freedom systems 631 The initial velocities in the normal coordinates are obviously zero. The solution of Equations r with the initial conditions (Equation v) gives y1 =

gt 2 mg + 2 k

) mg k cos 1.128 t y2 = −0.2886 k m ) mg k cos 2.175 t y3 = 0.2886 k m

(w)

Finally, the displacements in the physical coordinates are given by u = Qy or ! ) ) " mg gt 2 k k u1 = + 1 − 0.7854 cos 1.128 t − 0.2146 cos 2.175 t 2 k m m ! ) ) " mg gt 2 k k + 1 − 0.2886 cos 1.128 t + 0.2886 cos 2.175 t u2 = 2 k m m ! ) ) " mg gt 2 k k + 1 + 1.077 cos 1.128 t − 0.077 cos 2.175 t u3 = 2 k m m

(x)

SELECTED READINGS Biggs, J.M., Introduction to Structural Dynamics, McGraw-Hill, New York, 1964. Clough, R.W., and Penzien, J., Dynamics of Structures, 2nd Edition, McGraw-Hill, New York, 1993. Crandall, S.H., and McCalley, R.B. Jr., “Matrix Methods of Analysis’’, in Shock and Vibration Handbook, 4th Edition, (Ed. C.M. Harris), McGraw-Hill, New York, 1996. Hurty, W.C., and Rubinstein, M.F., Dynamics of Structures, Prentice-Hall, Englewood Cliffs, 1964. Meirovitch, L., Analytical Methods in Vibrations, Macmillan, London, 1967.

PROBLEMS 12.1

An eccentric mass shaker is mounted on the third-story level of the frame of Problem 10.6. The shaker has a rotating weight of 800 lb at an eccentricity of 12 in. with respect to the center of rotation. The speed of the rotor is 400 rpm. Obtain expressions for the steady state vibrations of the frame induced by the shaker.

12.2

Solve Problem 12.1 if the damping in each mode is 5% of critical.

12.3

The frame shown in Figure P8.15 is subjected to a lateral harmonic load p0 sin t at the top floor level, where p0 = 45 kN and  = 5π/3 rad/s. Derive expressions for the displacements at each floor level of the frame. Plot the top floor displacement as a

632

Dynamics of structures function of time for the first second of response. Find the maximum displacement of the top floor during this period and the time at which the maximum occurs. Neglect damping.

12.4

For Problem 12.3 plot the modal contributions to the top floor displacements as functions of time for the first second of response. For each of the three modes, determine the maximum response and the time at which it occurs during the first second. Can the absolute values of the maximum responses be added to obtain the maximum resultant response?

12.5

For Problem 12.3 plot the modal contributions to the base shear as well as the resultant base shear as functions of time for the first second of response. In each case determine the maximum value of the shear and the time at which it occurs.

12.6

For Problem 12.3 plot the steady state component of the top floor displacement in each of the three modes as a function of time for the first second of the response. What are the maximum values of the modal displacements and the resultant displacements? At what time do they occur?

12.7

The frame shown in Figure P8.15 is subjected to a rectangular pulse load of amplitude 50 kN and duration 0.5 s at the second floor level. Obtain expressions for the displacement at each floor level. Plot the top floor displacement as a function of time for the first second of response. Find the maximum response during this period and the time at which the maximum occurs. Neglect damping.

12.8

For Problem 12.7 plot the modal contributions to the top floor displacements as functions of time for the first second of response. For each of the three modes, determine the maximum response and the time at which it occurs during the first second.

12.9

For problem 12.7 plot the modal contributions to the interstory drift in second story for the first second of response. Also plot the resultant interstory drift. From this information find the maximum shear in the second story and the time at which it occurs.

12.10

A delicate instrument of mass m is packaged inside a rigid box of mass M (Fig. P12.10). The packaging material inside the box can be represented by a restraining spring of stiffness k and a viscous damper of coefficient c. The box is accidentally dropped from a height h. Derive expressions for the motion of the box and the instrument while in free fall.

Figure P12.10 12.11

Figure P12.11 shows a circular pipe section used in an industrial plant. The pipe is fixed to supports at a and b and has a 90◦ bend at c. It supports 2 heavy valves, each of mass 300 kg as indicated. The stiffness matrix of the pipe with respect to degrees of freedom 1 to 4 is given below. Find an expression for the total lateral displacement of

Forced dynamic response: Multi-degree-of-freedom systems 633

Figure P12.11 the valve at 1 due to the following ground motions (a) The support at a undergoes a harmonic motion along d.o.f. 3 given by ug = 10 sin 30t, where ug is the displacement in mm and t is the time in seconds; the support at b is fixed. (b) Both supports a and b have identical motion given by ug = 10 sin 30t. 

150.9 −41.1 EI  −41.1 150.9 K= 3 10.3 L −85.7 −65.1 30.9

−85.7 10.3 57.4 28.3

 −65.1 30.9  28.3 36.9

E = 200,000 MPa I = 0.5 × 106 mm4 12.12

The flexibility and stiffness matrices of the beam of Problem 11.1 are as given below   9 11 7 L3  11 16 11 A= 768EI 7 11 9   23 −22 9 768 EI  22 32 −22 K= 28 L3 9 −22 23 At t = 0 a constant load p0 is suddenly applied at A and remains in contact with the beam thereafter. Using the mode superposition method, obtain the displacements, shear forces and bending moments in the beam as functions of time. Neglect damping. If E = 200,000 MPa; I = 10.9 × 106 mm4 ; m = 900 kg, and L = 4 m, what is the bending moment under the load at t = 0.5 s? The load p0 is in kN.

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Chapter 13

Analysis of multi-degree-of-freedom systems: Approximate and numerical methods

13.1

INT RODUCTION

In characterizing the dynamic response of a system, the frequencies of vibration play a major role. The fundamental frequency of the system is often the quantity of primary interest. A great deal of attention has therefore been devoted to its determination. A particularly effective and simple method of estimating the fundamental frequency is the Rayleigh method described in Chapter 8. In this chapter we present an extension of Rayleigh’s method. Called the Rayleigh–Ritz method, it provides a means of improving the Rayleigh estimate for the fundamental frequency; in addition, it permits the determination of several higher frequencies of vibration. While the determination of one or more frequencies may be sufficient in some cases, in others, a complete history of response over a given period of time is required. Response history can be obtained by the mode superposition method of analysis, which was described in detail in Chapters 10 and 12. The method transforms the coupled equations of motion of an N-degree-of-freedom system into N independent singledegree-of-freedom equations. The response of the system is obtained by solving the N uncoupled equations and superposing the solutions. In the mode superposition method of analysis, if all N mode shapes of the system are used and the resulting uncoupled equations of motion are solved exactly, the exact response is obtained irrespective of the number of degrees of freedom the system may possess. In practice, even with modern powerful computers, the dynamic response analysis of a very large system with thousands of degrees of freedom is a very formidable and expensive task. In the mode superposition method, however, it is possible to obtain reasonable accuracy in the solution by including only a limited number of modes. How many modes need to be included in a particular case depends on the characteristics of the system and the applied load. In this chapter we discuss procedures for estimating the errors introduced by mode truncation and for determining the number of modes that must be included in the analysis so that a reasonable accuracy is obtained. The mode superposition method described in Chapters 10 and 12 is, in fact, a special case of the more general method based on the transformation of coordinates using appropriately selected Ritz shapes. Like mode superposition, the Ritz shape methods permit a reduction in the size of the problem. The generation of suitable Ritz vectors therefore plays an important role in the numerical analysis of the response

636

Dynamics of structures

of multi-degree-of-freedom systems. In this chapter we discuss several methods for the generation of such Ritz vectors. We also point out that many other methods of reducing the size of the problem, such as static condensation of the stiffness matrix and the mode acceleration method, belong to the category of Ritz vector methods. The Ritz vector methods lead to a reduced set of transformed equations of motion which must be solved in order to obtain the response of the original system. In special cases, for example when undamped vibration modes shapes are used as the Ritz vectors, and the system is either undamped or has proportional damping, the transformed equations are uncoupled and can be solved by an appropriate analytical or numerical method of solving single degree-of-freedom equations. Several such methods were described in Chapters 6 through 8. When the transformed equations are coupled, they may either be uncoupled by a second transformation or solved by a simultaneous direct integration of the entire set of reduced equations. The methods of integration that may be used in the latter case are similar to those described in Chapter 8 for the solution of single degree-of-freedom systems. In this chapter, we describe the application of several such methods to the numerical integration of multiple coupled equation. It should be noted that direct numerical integration can equally well be applied to the original system of equations of motion without any transformation.

13.2

RAYLEIGH–RITZ METHOD

Rayleigh’s principle and its applications have been discussed several times in previous chapters. It is instructive to summarize the different contexts in which the Rayleigh method was discussed. In Section 2.7.4 we described the application of an assumed shape function weighted by a generalized coordinate in the representation of a continuous system by a single-degree-of-freedom system. For example, by assuming that the displacement shape of a vibrating beam could be represented by u(x, t) = z(t)ψ(x), where z(t) was the generalized coordinate and ψ(x) a shape function, we derived the following equation of motion m∗ z¨ (t) + k∗ z(t) = p∗ where ∗



L

m =

2 m(x){ψ(x)} ¯ dx

0

∗



L

k =

EI(x){ψ  (x)}2 dx

0

and ∗



p =

L

pψ(x) dx 0

(13.1)

Analysis of multi-degree-of-freedom systems 637

For free vibrations, p∗ is zero and the solution of Equation 13.1 leads to the natural frequency of vibration ω. Thus ω2 =

k∗ m∗ 'L

(13.2a)

EI(x){ψ  (x)}2 dx = '0 L 2 dx ¯ 0 m(x){ψ(x)}

(13.2b)

Equation 13.2 was independently derived in Section 8.3.1 (Eq. 8.28) from energy considerations. As shown there, for free harmonic vibrations at the natural frequency, the maximum potential energy is given by Vmax = (U)max =

1 2 z 2



L

EI(x){ψ  (x)}2 dx

(13.3)

0

while the maximum kinetic energy is given by Tmax =

1 2 2 z ω 2



l

2 m(x){ψ(x)} ¯ dx

(13.4)

0

In the absence of any damping force, the energy must be conserved and hence Vmax must be equal to Tmax . The application of this condition leads to Equation 13.2b. The examples cited above are, in fact, applications of the Rayleigh method. The expression on the right-hand side of Equation 13.2b is called the Rayleigh quotient. Rayleigh’s principle states that this quotient takes its minimum value when ψ(x) is equal to the true vibration shape and that the minimum value of the quotient is equal to the square of the fundamental frequency. For any other value of ψ(x), the Rayleigh quotient is larger than the square of the fundamental frequency. Similar considerations apply to a multi-degree-of-freedom discrete parameter system. The application of a generalized coordinate and assumed vibration shape in the representation of such a system by an equivalent single-degree-of-freedom system was described in Section 2.8. As for a continuous system, the displacement vector in this case too is given by u = z(t)ψ, where ψ is now a vector. The equation of motion of the single-degree-of-freedom model is found to be m∗ z¨ + k∗ z = p∗ where m∗ = ψT Mψ k∗ = ψT Kψ and p∗ = ψT p

(13.5)

638

Dynamics of structures

An estimate for the fundamental frequency of the original system is obtained by solving Equation 13.5 with p∗ = 0. This gives k∗ m∗ ψT Kψ = T ψ Mψ

ω2 =

(13.6a) (13.6b)

Equation 13.6 could also be derived from energy considerations as illustrated in Example 8.3. The expression on the right-hand side of Equation 13.6b is the Rayleigh quotient and, as in the case of a continuous system, its minimum value obtained by varying ψ is equal to the square of the fundamental frequency. In Section 10.5 we presented a formal proof of the fact that Rayleigh quotient provides an upper bound estimate of the lowest eigenvalue or the square of the fundamental frequency. We further proved that if the arbitrary shape vector ψ is selected from a subset of vectors that are orthogonal to the first s − 1 eigenvectors, the Rayleigh quotient provides an upper bound estimate to the sth eigenvalue. Our discussions indicated that the quality of the estimate of fundamental frequency depended on the choice of shape function ψ: the closer this was to the true vibration mode shape, the better was the frequency estimate. In Section 8.4 we discussed a method of improving the frequency estimate. An extension to the Rayleigh method suggested by Ritz and known as the Rayleigh–Ritz method provides an alternative method of obtaining a better estimate of the fundamental frequency. At the same time, it can be used to obtain estimates of several higher frequencies. In the Ritz extension of Rayleigh method for a discrete parameter system, displacements of the system are expressed as a superposition of several different independent shape vectors, known as Ritz vectors, each weighted by its own generalized coordinate. Thus u = z1 (t)ψ1 + z2 (t)ψ2 + · · · + zM (t)ψM = z

(13.7)

where  is the matrix of Ritz vectors and z is the vector of M generalized coordinates. If we treat u as a possible displacement shape, the Rayleigh quotient corresponding to that shape is given by ρ= = =

uT Ku uT Mu zT  T Kz zT  T Mz ˜ zT Kz ˜ zT Mz

˜ =  T K and M ˜ =  T M. where K

(13.8)

Analysis of multi-degree-of-freedom systems 639

The value of Rayleigh quotient will change if the displacement shape is varied, which is equivalent to varying one or more of the generalized coordinates z1 , to zM . We know that in the vicinity of an eigenvalue of the system, ρ takes a stationary value. For a discrete system, the stationary value is a minimum near all eigenvalues except the highest, where it is a maximum. The conditions of stationarity can be stated as ∂ρ =0 ∂zj

j = 1, 2, . . . , M

(13.9)

For ease of reference, we denote the numerator of Equation 13.8 by v and the denominator by w. Note that both v and w are scalars. Equation 13.9 can now be expressed as 1 ∂v v ∂w − 2 =0 w ∂zj w ∂zj

j = 1, 2, . . . , M

(13.10a)

or as ˜ ∂ ∂ T˜ zT Kz ˜ =0 (z Kz) − (zT Mz) ˜ ∂zj ∂zj zT Mz

j = 1, 2, . . . , M

(13.10b)

˜ and M ˜ are symmetric, the complete set of M equations represented by Equation Since K 13.10b can be expressed as T ˜ ˜ − z Kz Mz ˜ =0 Kz ˜ zT Mz

(13.11)

Now recognizing that when the condition given by Equation 13.9 is satisfied, ρ takes the value ω2 , where ω is one of the frequencies of the system, we have ρ=

˜ zT Kz = ω2 ˜ zT Mz

(13.12)

Substitution of Equation 13.12 in Equation 13.11 gives ˜ = ω2 Mz ˜ Kz

(13.13)

Equation 13.13 will be recognized as a linearized eigenvalue problem. The frequencies ω obtained from its solution will be approximately equal to the frequencies of the original system, while the mode shapes z will be orthogonal to the reduced ˜ and M. ˜ Denoting a normalized value of z by φ, ˜ we have matrices K T ˜ φ˜ j = 0 φ˜ i K T ˜ φ˜ j = 0 φ˜ i M

i = j

where φ˜ i is the ith normalized mode shapes.

(13.14)

640

Dynamics of structures

Mode shapes φ˜ are of size M and are not the mode shapes of the original system, which is of size N. It is, however, possible to obtain from φ˜ i the mode shapes of the original system. To do this, we apply the transformation given by Equation 13.7: φi =  φ˜ i

i = 1, 2, . . . , M

(13.15a)

or ˜  = 

(13.15b)

Vectors φ are approximations to M mode shapes of the original system and are orthogonal with respect to both the K and M matrices. To demonstrate this, we note that T φTi Kφj = φ˜ i  T K φ˜ j T ˜ φ˜ j = φ˜ i K

i = j

(13.16)

=0 where the last step follows from Equation 13.14. In a similar manner, T ˜ φ˜ j φTi Mφj = φ˜ i M =0

i = j

(13.17)

Also, T ˜ φ˜ i φ˜ i K φTi Kφi = T T φi Mφi ˜ φ˜ i φ˜ i M

(13.18)

= ωi2 Again the last relationship holds because φ˜ i and ωi are the eigenpairs obtained from the solution of Equation 13.13. The frequencies and mode shapes obtained as above by application of the Rayleigh–Ritz method are only approximations to the true frequencies and mode shapes. The quality of the approximation depends on the selection of the assumed shapes ; the closer these shapes are to the true vibration mode shapes, the better are our estimates of the system frequencies and mode shapes. As proved in Section 10.5 (Eq. 10.38), the frequency estimates will always be larger than the fundamental frequency and smaller than the highest frequency. The Rayleigh Ritz method will also provide estimates for intermediate frequencies, but what particular frequencies are estimated will depend on the selection of the Ritz shapes. In addition, the estimate of an intermediate frequency may be smaller or larger than the closest true frequency. For a complex multi-degree-of-freedom system, the selection of vibration shapes to be used in Rayleigh Ritz method is not simple. It is, however, possible to improve an initial estimate of the vibration shape by following a procedure very similar to that used in the Rayleigh method for obtaining the fundamental frequency of a multi-degree-offreedom system.

Analysis of multi-degree-of-freedom systems 641

In the improvement to the Rayleigh method described in Section 8.4, we reasoned that the deflected shape obtained by the application of inertia forces resulting from an initial assumption of the free-vibration shape will provide a better estimate of the true free-vibration shape. Similar reasoning can be applied in application of the Rayleigh– Ritz method. Thus, if the initial assumption of the vibration shapes is  (0) , the deflected shapes under the action of inertia forces resulting from  (0) will be given by an equation parallel to Equation 8.36: 

(1)

= ω2 K−1 M (0)

(13.19)

Since ω2 is not known, we use vibration shapes  (1) that are proportional to  and are obtained by dropping ω2 from Equation 13.19. Thus  (1) = K−1 M (0)

(1)

(13.20)

and 

(1)

= ω2  (1)

(13.21)

Use of  (1) in place of  in Equation 13.7 leads to the improved Rayleigh–Ritz equations ˜ (1) z ˜ (1) z = ω2 M K

(13.22)

where ˜ (1) =  (1) T K (1) K T

=  (0) MK−1 M (0) T

=  (0) MFM (0)

(13.23)

and ˜ (1) =  (1) T M (1) M T

=  (0) MK−1 MK−1 M (0) T

=  (0) MFMFM (0)

(13.24)

in which F is the flexibility matrix. The procedure outlined by Equations 13.22, 13.23, and 13.24 permits improvements in the calculated vibration shapes so that reasonable estimates of frequencies can be obtained from crude initial assumption of the vibration shapes. In addition, it allows use of the flexibility matrix rather than the stiffness matrix in the computations, which may be an advantage in certain cases. An explicit evaluation of the flexibility matrix is not, in fact, needed; calculation of the deflections caused by applied inertia loads equal to M (0) is all that is required. As a final comment, the procedure will be recognized as the first step in the subspace iteration method described in Section 11.5.4.

642

Dynamics of structures

Example 13.1 Obtain the stiffness and mass matrices for the five-story building frame shown in Figure E13.1a. Using the Rayleigh–Ritz method and the following displacement shapes, obtain two frequencies and the corresponding mode shapes of the building. Then by using an improved Rayleigh–Ritz method, obtain better estimates of the frequencies and mode shapes. 

 (0)

0.2 0.4  = 0.6 0.8 1.0

 −0.5 −1.0  −0.5  0.5 1.0

Solution Corresponding to the degrees of freedom shown in Figure E13.1a, the mass and stiffness matrices of the frame are given by 



2

  M=  

    

2 2 2 1



2800 −1200  0 K=    0 0

−1200 2400 −1200 0 0

0 −1200 2000 −800 0

0 0 −800 1200 −400

 0 0  0  −400 400

Figure E13.1 (a) Five-story building frame; (b) displacements caused by forces Mq1 ; (c) displacements caused by forces Mq2 .

Analysis of multi-degree-of-freedom systems 643 The transformed mass and stiffness matrices are obtained as ˜ (0) = ( 0 )T M 0 M   3.4 0.2 = 0.2 4.5

(a)

˜ (0) = ( 0 )T K 0 K   208 40 = 40 1900

(b)

The reduced eigenvalue problem of Equation 13.13 becomes  208 40

 0.2 z=0 4.5

  3.4 40 z−λ 0.2 1900

(c)

The eigenvalues can be obtained by solving the characteristic equation of the eigenproblem. Thus ( ( (208 − 3.4λ 40 − 0.2λ ( (=0 ( (d) ( 40 − 0.2λ 1900 − 4.5λ( or λ2 − 483.6λ + 25,790 = 0

(e)

On solving Equation e we get λ1 = 61.04 λ2 = 422.6

ω1 = 7.81 ω2 = 20.56

(f)

Substitution of λ1 and λ2 , in turn, in Equation c give the corresponding eigenvectors in the reduced space   −58.481 1   −0.03623 = 1

z1(1) = z2(1)

(g)

The eigenvectors of the original system are obtained from q(1) =  (0) z(1)

(h)

Substitution for z from Equation g in Equation h gives  −12.196 −24.392    = −35.588 −46.285 −52.481 

q1(1)

 −0.5072 −1.0145    = −0.5217  0.4710 0.9638 

q2(1)

(i)

644

Dynamics of structures

When normalized so that the last element of the vector is 1, the eigenvectors q1 and q2 become     0.212 −0.5263 0.424 −1.0526     (1) (1)    (j) q1 = 0.619 q2 =  −0.5414 0.805  0.4887 1.000 1.000 For the purpose of comparison, the exact values of the first two eigenvalues and eigenvectors are given below. λ1 = 60.39 λ2 = 360.4

ω1 = 7.77 ω2 = 18.98

 0.199 0.445    q1 =  0.645 0.849 1.000 

 −0.309 −0.536    q2 =  −0.440  0.995 1.000 

(k)

The third eigenvalue is λ3 = 765.08 or ω3 = 22.66. We have obtained a good approximation to the first frequency. The second frequency estimate lies between the second and third true frequencies, but is closer to the second true frequency. The second frequency estimate is not as good as the first; and the estimate for the second eigenvector is, in fact, quite poor. To obtain an improved estimate of the eigenvalues and eigenvectors, we subject the frame to inertia forces caused by vibrations in each of the two mode shapes just derived. These inertia forces are proportional to Mq1 and Mq2 , respectively. Next, we obtain the displacements caused by these inertia forces. The computations for obtaining the displaced shapes are straightforward and are shown in Figure E13.1b and c. The calculations in Figure E13.1b and c are self-explanatory. Since we are interested only in a displacement shape rather than in the absolute value of the displacements, we have normalized the displacements so that the deflection at the top-story level is 1. This keeps the numbers within reasonable bounds. The resulting displaced shapes are  0.1990 0.4423    = 0.6417 0.8456 1.0000 

q1(1)

 −0.4296 −0.7356    = −0.5094  0.2410 1.0000 

q2(1)

Next we use the displacement shapes just obtained as the new Ritz vectors to transform the problem to that of a reduced size. This gives ˜ (1) =  (1) T M (1) M   3.7241 −0.0679 = −0.0679 3.0865 (1) (1) T (1) ˜ K =  K   225.13 −3.21 = −3.21 1149.97 in which we have used  (1) = [q1(1) q2(1) ], q1(1) and q2(1) being the latest values of Ritz vectors.

(l)

Analysis of multi-degree-of-freedom systems 645 The characteristic equation of the reduced eigenvalue problem becomes λ2 − 433.12λ + 22,510 = 0

(o)

Solution of Equation m leads to the following eigenvalues λ1 = 60.393 λ2 = 372.73

ω1 = 7.771 ω2 = 19.306

(n)

Substitution of λ1 and λ2 , in turn, in the eigenvalue equation gives the following eigenvectors in the reduced space   −1009 1   0.019438 = 1

z1(2) = z2(2)

(o)

The eigenvectors of the original system are obtained from q(2) =  (1) z(2)

(p)

After normalization the new eigenvectors are given by  0.1996 0.4435    q1 =  0.6443 0.8461 1.0000 

 −0.4174 −0.7131    q2 =  −0.4874  0.2526 1.0000 

(p)

The frequency estimates have improved for both the first and second modes.

Example 13.2 Obtain initial estimates of two of the frequencies of the system of Example 13.1 by using the following assumed vibration shapes in the Rayleigh–Ritz method 

1 4 1 3  1 3

   ψ1 =      0 − 34



1 3 1 3

     ψ2 =   0  1 − 2  1 2

Solution The transformed mass and stiffness matrices are given by ˜ =  T M M  1.3190 = 0.0139

 0.0139 1.1944

(a)

646

Dynamics of structures ˜ =  T K K 

1.0566 = 400 −0.0833

 −0.0833 2.7778

(b)

The reduced eigenvalue problem becomes   1.1319 −0.0833 z−λ 0.0139 2.7778



1.0566 400 −0.0833

 0.0139 z=0 1.1944

(c)

On setting λ = λ/400, the characteristic equation of the eigenvalue problem can be written as 2

λ − 2.8426λ + 1.7753 = 0

(d)

Solution of Equation d gives λ1 = 0.9255 λ2 = 1.9161

λ1 = 370.22 λ2 = 766.43

ω˜ 1 = 19.24 ω˜ 2 = 27.68

The exact values of the first four frequencies are ω1 = 7.77, ω2 = 18.98, ω3 = 27.66, and ω4 = 36.39. The estimates that we have obtained, ω˜ 1 and ω˜ 2 , are closest to the true frequencies ω2 and ω3 . It is apparent from Examples 13.1 and 13.2 that the eigenvalue estimates obtained by the Rayleigh–Ritz are greatly influenced by the selection of Ritz shapes.

Example 13.3 The two starting Ritz vectors used to obtain frequency estimates for the system of Example 13.1 happen to satisfy the following relationships ψ1 = aφ1 + bφ2 ψ2 = cφ3 + dφ4 where a, b, c, and d are arbitrary constants. (a)

(b)

Prove that the first frequency estimate will lie between the first and the second true frequencies, while the second frequency estimate will lie between the third and the fourth true frequencies. On selecting a = 0.2, b = 0.9, c = 0.9, and d = 0.2, the following trial Ritz vectors are derived  0.20  0.32    ψ1 =   0.23 −0.18 −0.84 

 0.18  0.31    ψ2 =   0.06 −0.61 0.61 

in which the results have been rounded off to two decimal digits. Using these trial vectors obtain two frequency estimates.

Analysis of multi-degree-of-freedom systems 647

Solution (a) Using the orthogonality relationships the transformed mass and stiffness matrices can be shown to be   2 2 0 ˜ = a +b (a) M 2 2 0 c +d  2 2 2 2 ˜ = a ω1 + b ω2 K 0

0 c2 ω32 + d 2 ω42

 (b)

Because the reduced mass and stiffness matrices are both diagonal, the two frequencies of the reduced problem are obtained quite readily and are given by ω˜ 12 =

a2 ω12 + b2 ω22 a2 + b 2

(c)

ω˜ 22 =

c2 ω32 + b2 ω42 c2 + d 2

(d)

Equation c can be expressed as ω˜ 12 =

a2 (ω12 /ω22 ) + b2 2 ω2 a2 + b 2

(e)

Since ω1 is smaller than ω2 , the numerator in Equation e is smaller than the denominator. As a result ω˜ 1 is smaller than ω2 . As an alternative, Equation c can be expressed as ω˜ 12 =

a2 + b2 (ω22 /ω12 ) 2 ω1 a2 + b 2

(f)

Again, because ω2 is larger than ω1 the numerator in Equation f is larger than the denominator and hence ω˜ 1 is larger than ω1 . Thus the first frequency estimate lies between the true first frequency and the true second frequency. In a similar manner, we can prove that the second frequency estimate ω˜2 will lie between the true third and fourth frequencies. (b) For the supplied Ritz vectors the transformed mass and stiffness matrices work out to  1.1610 ˜ M= 0.0052  ˜ = 399.7 K 1.0

 0.0052 1.3805

(g)



1.0 1101.6

(h)

which are very nearly diagonal. The solution of the reduced eigenvalue problem provides the following two estimates. ω˜1 = 18.56

ω˜2 = 28.25

(i)

The first estimate is very close to the second true frequency of 18.98, but is smaller than the true frequency. The second estimate is close to the the third true frequency of 27.66, but is larger than the true frequency. It is evident that the Rayleigh–Ritz method may provide estimates for intermediate frequencies, but which true frequencies are matched depends on the selection of the starting Ritz vectors. Also, the intermediate frequency estimates may be higher or lower than the

648

Dynamics of structures

closest true frequencies. In our example the first starting vector was close to the second mode shape, hence it led to a good frequency estimate for the second frequency. However, because of contamination from the first mode shape, the estimated frequency was lower than the true frequency. It may be noted that as proved in Section 10.5 a frequency estimate provides an upper bound to a true frequency only when the Ritz shape used to obtain the frequency estimate is orthogonal to all mode shapes lower than the one for which the frequency is being estimated. In our example the second trial shape is orthogonal to all modes lower than 3, as a result the estimate obtained for the true third frequency is an upper bound.

The Rayleigh–Ritz method is also applicable to continuous systems. Similar to Equation 13.7, the deflected shape of a continuous system can be represented by a superposition of M shape functions each weighted by a different generalized coordinate. As an example, for the lateral vibrations of a beam, the deflected shape is represented by u(x, t) = z1 (t)ψ1 (x) + z2 (t)ψ2 (x) + · · · + zM (t)ψM (x) = ψz

(13.25)

where ψi , i = 1 to M, are Ritz shapes which are functions of the spatial coordinates x; zi , i = 1 to M, are the generalized coordinates; ψ is a row vector of shapes ψi ; and z is the vector of generalized coordinates. The Rayleigh quotient corresponding to the displaced shape u is given by 'L

EI(x)zT [ψ (x)]T ψ (x)z dx ρ = 0' L T [ψ(x)]T ψ(x)z dx ¯ 0 m(x)z

(13.26)

We now introduce the notations ˜ = K ˜ = M

 

L

EI(x)[ψ (x)]T ψ (x) dx

(13.27a)

T m(x)[ψ(x)] ¯ ψ(x) dx

(13.27b)

0 L 0

so that k˜ ij =

 

L 0 L

m ˜ ij =

EI(x)ψi (x)ψj (x) dx

(13.28a)

m(x)ψ ¯ i (x)ψj (x) dx

(13.28b)

0

With the notations given by Equation 13.27, Equation 13.26 reduces to the form of Equation 13.8. Application of stationarity condition on the Rayleigh quotient then leads to the eigenvalue equation (Eq. 13.13).

Analysis of multi-degree-of-freedom systems 649

Example 13.4 For the lateral vibrations of uniform cantilever beam shown in Figure E13.4a, determine two frequencies and mode shapes using the shape functions given below and shown in Figure E13.4b. πx 2L 3πx ψ2 = 1 − cos 2L ψ1 = 1 − cos

Both shape functions satisfy the two geometric conditions at x = 0. They also satisfy the zero moment condition at x = L but not the zero shear condition at the free end of the cantilever.

Figure E13.4 (a) Uniform cantilever beam; (b) vibration shape functions; (c) mode shapes for the lateral vibrations of uniform cantilever.

650

Dynamics of structures

Solution The eigenvalue problem is given by Equation 13.13: ˜ = ω2 Mz ˜ Kz

(a)

in which the elements of the stiffness and mass matrices are obtained from Equations 13.28a and 13.28b, respectively. We thus have k11 =

1 π 4 EI 16 L4

1 π 4 EI = 32 L3 k12 =

9 π 4 EI 16 L4

= k21 k22

81 π 4 EI = 16 L4 =



L



cos

0



L

cos 0

 0

L




πx 2 dx 2L

3πx πx cos dx = 0 2L 2L

3πx cos 2L

(b)

2 dx

81 π 4 EI 32 L3

Also, 

 πx 2 m ¯ 1 − cos dx 2L 0 
3L 4L − =m ¯ 2 π
  L  πx  3πx = m ¯ 1 − cos 1 − cos dx 2L 2L 0 
4L =m ¯ L− 3π

m11 =

m12

m22

L

(c)

= m21   L
3πx 2 = m ¯ 1 − cos dx 2L 0 
4L 3L + =m ¯ 2 3π

Substituting Equations b and c in Equation a, we get  1 1 π 4 EI  3 32 L 0

  3 z1 − π4 2 2    = ω mL ¯  4 81 z2 1 − 3π 0

  z1   4 z2 + 3π

1− 3 2

4 3π

(d)

On setting λ¯ = (32mL ¯ 4 /π 4 EI)ω2 , we get the following characteristic equation from Equation d: ( (1 − 0.2268λ¯ ( ( ( ( −0.5756λ¯

( −0.5756λ¯ (( (=0 ( 81 − 1.9244λ¯ (

(e)

Analysis of multi-degree-of-freedom systems 651 or λ¯ 2 − 192.948λ¯ + 770.074 = 0

(f)

Solution of Equation f gives )

λ¯ 1 = 4.0775 λ¯ 2 = 188.87

EI mL ¯ 4 ) EI ω2 = 23.978 mL ¯ 4

ω1 = 3.523

The exact values of the first two frequencies are ) EI ω1 = 3.516 mL ¯ 4 ) EI ω2 = 22.034 mL ¯ 4

(g)

(h)

The approximate frequencies are fairly close to the exact values of the first two frequencies. Also, as expected, the approximate values provide upper bound estimates of the true frequencies. Substitution of λ¯ 1 and λ¯ 2 , in turn, in Equation d provides the following values for the generalized coordinates     z1 1 = 0.0321 z2     z1 1 = −0.3848 z2

(i)

The mode shapes are now obtained from Equation 13.25 as
  3πx πx  q1 = 1 − cos + 0.0321 1 − cos 2L 2L 3πx πx − 0.0321 cos = 1.0321 − cos 2L 2L
   πx 3πx − 0.3848 1 − cos q2 = 1 − cos 2L 2L 3πx πx + 0.3848 cos = 0.6152 − cos 2L 2L

(i)

The two mode shapes have been plotted in Figure E13.4c.

Example 13.5 Assuming that the cantilever beam of Example 13.4 is of nonuniform section so that its moment of inertia and mass vary as follows, obtain estimates of two frequencies of the beam.  x  m(x) ¯ =m ¯0 1− 2L  x  I(x) = I0 1 − 2L

652

Dynamics of structures

Solution The elements of stiffness and mass matrices are again obtained from Equations 13.28a and 13.28b, respectively. Thus 

x  1 π 4 EI0 2L 16 L4 0
 1 π 4 EI0 3 1 = + 16 L3 8 2π 2  L x  9 π 4 EI0 = 1− 2L 16 L4 0 
4 9 π EI0 1 = 2π 2 16 L3 L

k11 =

k12

k22



1−



πx 2 dx 2L




 3πx πx  cos dx 2L 2L

cos

cos

(a)

= k21 
 L x  81 π 4 EI0 3πx 2 = dx 1− cos 2L 16 L4 2L 0
 81 π 4 EI0 3 1 = + 16 L3 8 18π 2

Also, 

 x  1 − cos m ¯0 1− 2L 0
 7L 9L 2L − − =m ¯0 8 π 2π 2  L  x  1 − cos = m ¯0 1− 2L 0
 2L 31L 3L − − =m ¯0 4 3π 18π 2

m11 =

m12

m22

L

πx 2 dx 2L


 πx  3πx 1 − cos dx 2L 2L (b)

= m21
  L  3πx 2 x  1 − cos = m ¯0 1− dx 2L 2L 0
 2L 7L 9L + − =m ¯0 8 3π 18π 2

The eigenvalue equation becomes π 4 EI0 16L3

3 8

+

1 2π 2

9 2π 2

81

3 8

  z1

9 2π 2

+

1 18π 2



z2

¯ 0L =ω m 2

9 8 3 4

−

−

2 π

−

2 3π

−

7 2π 2 31 18π 2

3 4

−

2 3π

−

9 8

+

2 3π

−

31 18π 2 7 18π 2

  z1 z2

or π 4 EI0 0.4257 16L3 0.4559

  0.4559 z1 30.8310

z2

¯ 0L =ω m 2

0.1337

0.3633

0.3633

1.2980

  z1 z2

(c)

Analysis of multi-degree-of-freedom systems 653 On setting λ¯ = (16m ¯ 0 L4 /π 4 EI0 )ω2 , Equation c leads to the following characteristic equation: ( (0.4257 − 0.1337λ¯ ( ( (0.4559 − 0.3633λ¯

( 0.4559 − 0.3633λ¯ (( (=0 30.8310 − 1.2980λ¯ (

or λ¯ 2 − 104.63λ¯ + 311.176 = 0 Solution of Equation d gives the two frequencies ) EI0 λ¯ 1 = 3.0638 ω1 = 4.319 mL ¯ 4 ) EI0 λ¯ 2 = 101.57 ω2 = 24.87 mL ¯ 4

(d)

(e)

The discussion in the foregoing paragraphs and Examples 13.1 through 13.5 demonstrate that the reliability of frequency estimates obtained from the Rayleigh– Ritz method depends on a proper selection of the Ritz shapes. Unless there is some indication of the nature of the vibration shape of the system in the desired mode, the frequency estimates can be very inaccurate. In general, the Rayleigh–Ritz method is used to obtain the lowest few frequencies of the system. The selected Ritz shapes should therefore resemble the lowest modes. For complicated systems, it is extremely difficult to construct such shapes. Despite these shortcomings, the Rayleigh–Ritz method can be quite useful for estimating the lowest few frequencies whenever good Ritz shapes can be derived on the basis of experience and judgment. The method is also very useful for continuous systems, particularly for continuous systems having nonuniform properties where the exact eigenvalues are impossible to determine.

13.3 APPLICATION OF RITZ METHOD TO FORCED VIBRATION RESPONSE The Rayleigh–Ritz method, in fact, represents a transformation of coordinates in which the transformation matrix consists of a series of linearly independent vectors called Ritz vectors or Ritz shapes. If the number of such vectors is equal to the number of degrees of freedom in the system being analyzed, the solution of the eigenproblem in the transformed space leads to the exact eigenvalues and eigenvectors of the original system. The value of the Rayleigh–Ritz method, however, lies in the fact that with appropriate choice of Ritz vectors, only a few of them may be used to represent the original system adequately. The transformation matrix is, in such a case, rectangular and the size of the transformed problem is considerably smaller than the original problem. Solution of the reduced problem then provides approximations to the true eigenvalues and eigenvectors of the original system. The position of the eigenvalues that will be approximated and the accuracy with which they are calculated depends on the choice of the Ritz vectors. The fact that with appropriate selection of the Ritz shapes, only a few of them can be used to represent adequately the response of the original system can be utilized

654

Dynamics of structures

very effectively in the forced response analysis of multi degree-of-freedom systems. The use of Ritz method leads to a significant reduction in the size of the problem and therefore in the computations involved in the solution. However, because the quality of the results obtained is influenced strongly by the selection of Ritz shapes, methods for selecting such shapes are of considerable importance. In this section we discuss several different approaches used in the selection of Ritz shapes, the details of the solution procedure in each case, and where applicable, the errors involved in the solution.

13.3.1 Mode superposition method The mode superposition method of solution of the forced vibration response, discussed in Chapter 12, will readily be recognized as a Ritz method in which the undamped mode shapes of the system are used as the Ritz shapes. It was shown that the application of the method leads to N uncoupled single-degree-of-freedom equations given by y¨ n + 2ξn ωn y˙ n + ωn2 yn = φTn p

n = 1, 2, . . . , N

(13.29)

in which it is assumed that mass-orthonormal mode shapes have been used as the Ritz vectors and that the damping is proportional, so that the transformation diagonalizes the damping matrix. Once the single-degree-of-freedom equations given by Equations 13.29 have been solved, the response in the physical coordinates is obtained from u=

M

φn yn

(13.30)

n=1

in which we have superimposed the contributions from only the the first M mode shapes. This implies that we need to evaluate only the first M eigenpairs and further that only the first M equations from the N equations represented by Equation 13.29 need be solved. Obviously, an important consideration is how many mode shapes should be included in the computations so as to obtain reasonable accuracy. To find an answer to this question, let us first assume that the forcing function is of the form p = f sin t, where f is a time-independent amplitude vector and sin t represents the time variation. The steady-state solution to nth equation in Equations 13.29 is then given by (Eq. 6.27) yn =

1 φTn f sin(t − θ)  ωn2 (1 − βn2 )2 + (2ξn βn )2

(13.31)

in which βn = /ωn and tan θ = 2ξn βn /(1 − βn2 ). It is obvious that the characteristics of the forcing function that affect the response are its amplitude vector f and the frequency . The effect of amplitude is represented by the term γ = φTn f

(13.32)

which is called the participation fator. The effect of the exciting frequency  is reflected  in the term AD = 1/ (1 − βn2 )2 + (2ξn βn )2 . This term is plotted in Figure 6.4 as a function of βn for several values of the damping fraction ξn . For small values of βn , that is,

Analysis of multi-degree-of-freedom systems 655

when ωn is much larger than , say four times  or more, AD is very nearly equal to 1. Also, as seen in Figure 6.5, the phase angle θ tends to zero as β approaches zero. Hence the modal response yn reduces to yn = =

φTn f sin t ωn2 φTn p kn